# Linear-semiorders and their incomparability graphs

A linear-interval order is the intersection of a linear order and an interval order. For this class of orders, several structural results have been shown. In this paper, we study a natural subclass of linear-interval orders. We call a partial order a linear-semiorder if it is the intersection of a linear order and a semiorder. We show a characterization of linear-semiorders in terms of linear extensions. This gives a vertex ordering characterization of their incomparability graphs. We also show that being a linear-semiorder is a comparability invariant.

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• 3 publications
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## Preliminaries

A partially ordered set is a pair , where is a set and is a binary relation on that is irreflexive, transitive, and therefore asymmetric. The set is called the ground set while the relation is called a partial order on . In this paper, we will deal only with partial orders on finite sets. We denote partial orders by instead of , that is, we write in if and only if . Two elements are comparable in if or ; otherwise and are incomparable, which we denote . A partial order on a set is a linear order if any two elements of are comparable in . A partial order on a set is an interval order if for each element , there is a (closed) interval on the real line so that for any two elements , we have in if and only if lies completely to the left of . Here, the interval lies completely to the left of , and we write , if . The set of intervals is called an interval representation of . An interval representation is unit if every interval has unit length, and it is proper if no interval properly contains another. An interval order is a semiorder if it has a unit interval representation. It is known that a partial order is a semiorder if and only if it has a proper interval representation [BW99-DM]. Let and be two partial orders on the same ground set . The intersection of and is the partial order . Equivalently, the intersection of and is the partial order on such that in if and only if in both and . We call an order a linear-semiorder if it is the intersection of a linear order and a semiorder. Let be a partial order on a set . The comparability graph of is the graph such that if and only if and are comparable in ; the incomparability graph of is the graph such that if and only if in . A cocomparability graph is the complement of a comparability graph. Note that any cocomparability graph is the incomparability graph of some partial order.

## Comparability invariance

A property of partial orders is a comparability invariant if either all orders with the same comparability graph have that property or none have that property. It is known that being a linear-interval order is a comparability invariant [COS08-ENDM]. In this section, we will show the following. Being a linear-semiorder is a comparability invariant. We use the proof technique developed in [FM98-Order]. Let be a partial order on a set . A non-empty subset is autonomous in if for any element , whenever , , or holds in for some element , then the same holds for all elements . Let be a partial order having the same comparability graph as . The order is obtained by a reversal from if there is an autonomous set of such that:

1. If not both of and are in , then in if and only if in .

2. If both of and are in , then in if and only if in .

We denote by the order obtained from by reversing . The following theorem [Gallai67] provides a simple scheme to show the comparability invariance results: Two orders and have the same comparability graph if and only if there is a finite sequence of orders such that , , and is obtained from by a reversal for each . Therefore, in order to prove Theorem Document, we will show the following claim. If an order on a set is a linear-semiorder and a subset is autonomous in , then is a linear-semiorder. Recall that and are two horizontal lines in the -plane with above . As a representation of a linear-semiorder, we use a set of triangles between and as follows. Let be a linear-semiorder on a set , and let and be a linear order and a semiorder on with . Let be a set of points on such that if and only if in for any two elements . Notice that all the points are distinct by definition. Let be a unit interval representation of on . We assume that no two intervals share a common endpoint. Let for each element . Since each interval has unit length, . Let be the triangle spanned by and . A triangle lies completely to the left of , and we write , if on and on . We have that in if and only if for any two elements ; hence we call the set a triangle representation of . Note that in the following, we use the term triangle to denote a triangle spanned by a point on and a unit-length interval on . Now, we start to prove Theorem Document. We fix a pair of a linear-semiorder and an autonomous set of . We also fix a triangle representation of . An element is isolated if in for any element . Let be the subset of obtained by removing all isolated elements of . We can observe the following. The set is autonomous in , and . Thus we assume without loss of generality . We define as the convex region spanned by the triangles with . We also define that and , similarly, and . Notice that is the trapezoid spanned by the interval on and the interval on , while is the upper left corner of and is the upper right corner of , similarly, is the lower left corner of and is the lower right corner of . Moreover, let and denote the elements of with and , respectively. Similarly, let and denote the elements of with and , respectively. Let be the set of elements with . Obviously, . The convex region spanned by the triangles with is equal to . Let . If then intersects with every triangle contained in . Note that the triangles in the lemma are not just the triangles of the representation of , but every triangle spanned by a point on with and a unit-length interval on with . Since , we have , and hence or . Suppose . Since , we have . Thus , and hence in . Since is autonomous, in , and hence . Thus . Therefore, we have and , and the claim holds. A similar argument would show that the claim holds when . Suppose . If then . If then implies , and hence . Thus in both cases, in for some element . Since is autonomous, in . Since is not isolated, there is an element with . Since and each interval has unit length, , and hence . It follows that . Therefore, we have and , and the claim holds. A similar argument would show that the claim holds when . As a consequence of Lemma Document, we have the key lemma. For any element , one of the following holds:

1. [label=–]

2. lies completely to the left of .

3. intersects with every triangle contained in .

4. lies completely to the right of .

We also have the following from Lemma Document The set is autonomous in . Let . Suppose in for some element . Then implies , and hence for all elements . Thus in for all elements . If in for some element , then lies completely to the left of , and hence in for all elements . If in for some element , then lies completely to the right of , and hence in for all elements . Let . We define as the set of triangles obtained from the representation of by flipping the representation of relative to , that is, for any element while for each element , the triangle is the triangle spanned by the point and the interval . The convex region spanned by the triangles with is equal to . The order is a linear-semiorder with a representation . Let be the partial order on defined by the representation . If neither nor is in , then their relation in remains the same as in . If one of and is in , then Lemma Document ensures that their relation in remains the same as in . If both and are in , then ; hence in if and only if in . Thus . Let . If then . It follows that is a linear-semiorder. Thus we assume without loss of generality . All the elements of is incomparable to all the elements of in . Suppose that there is an element with in for some element . Since is autonomous, and in , contradicting to . Similarly, we have that there is no elements with in for some element . The set is autonomous in , and . Let . If then, since is autonomous in , whenever , , or holds in for some element , then the same holds for all elements . If then , and we have from Lemma Document that in for all elements . Thus is autonomous in . Lemma Document implies . Thus . Since and , we have . We repeat the process replacing and with and . Let be the subset of obtained by removing all isolated elements of . By Lemma Document, the set is autonomous in and . We assume without loss of generality . We define as the convex region spanned by the triangles with . Let be the set of elements with . By Lemma Document, the set is autonomous in , and let . By Lemma Document, the order is a linear-semiorder. Let , and we assume without loss of generality . By Lemma Document, the set is autonomous in , and . Let be the subset of obtained by removing all isolated elements of . By Lemma Document, the set is autonomous in , and . The set is a proper subset of . Since , we have , and hence . We also have that the inclusion is proper since, for example, . Recall that can be partitioned into three sets , , and . By definition, all the elements are isolated in . By Lemma Document, all the elements of are incomparable to all the elements of . Thus they are isolated in . We have . Since all the elements of are isolated in , we have and . Since but , the inclusion is proper. Now, we are ready to prove the main theorem. [Proof of Theorem Document] Suppose that the theorem does not hold. It follows that there is a pair of a linear-semiorder and an autonomous set of such that is not a linear-semiorder. Among such pairs, we choose one such that is minimal. We have from Lemma Document that , but we also have from Lemma Document that , contradicting to the minimality of .

## Characterization

Let be a partial order on a set . A linear order on is a linear extension of if in whenever in . Hence, the linear extension of has all the relations of with the additional relations that make linear. We define some properties of linear extensions. The order of is the partial order consisting of four elements , , , of such that and while and in . Notice that and in ; for otherwise we would have or in . We say that a linear extension of fulfills the rule if or in for each induced suborder of ; if in then in ; if in then in . Equivalently, a linear extension of is said to fulfill the rule if there is no four elements of such that , , , and in while and in . See Fig. Documentfig:forbidden:2+2. We call such an induced suborder a forbidden configuration for . The order of is the partial order consisting of four elements , , , of such that while and in . Notice that in ; for otherwise we would have or in . We say that a linear extension of fulfills the rule if or in for each induced suborder of ; if in then in ; if in then in . Equivalently, a linear extension of is said to fulfill the rule if there is no four elements of such that , , and in while in . See Fig. Documentfig:forbidden:3+1. We call such an induced suborder a forbidden configuration for .

Our previous work [Takaoka18-DM] shows that a partial order is a linear-interval order if and only if it has a linear extension fulfilling the rule. In this paper, we show a similar characterization for linear-semiorders. A partial order is a linear-semiorder if and only if it has a linear extension fulfilling the rule and rule. The necessity and sufficiency follow immediately from Lemmas Document and Document, respectively. If a partial order on a set has a linear order and a semiorder with , then has no forbidden configurations for nor . It is shown in [CK87-CN, Takaoka18-DM] that has no forbidden configurations for . Suppose for a contradiction that has a forbidden configuration for consisting of four elements of such that , , and in while in . Let be a proper interval representation of , and let for each element of . Since in , we have . Since in and in , we have in . Thus . Similarly, since in and in , we have in . Thus . Therefore, we have , contradicting to that no interval properly contains another. If a partial order on a set has a linear extension fulfilling the rule and rule, then there is a semiorder with . The semiorder and its proper interval representation can be obtained in time, where is the number of elements of

. In this proof, we often denote partial orders in the formal sense, that is, as a set of ordered pairs of elements. For example, we write

if in ; we write if in and in . Suppose that there is such a semiorder . Let be a proper interval representation of , and let for each element of . Let and be two elements of . Suppose that there is an element with and . Then and hence . Since implies , we have . Similarly, if there is an element with and , then , and hence . Trivially, if . To capture these relations, we define the following. Let be the binary relation on such that if there is an element with and ; let be the binary relation on such that if there is an element with and . Let . Note that the binary relation is not transitive in general. See for example the partial order in Fig. Documentfig:intransitivity; we have and , but . There is a linear order on such that in whenever . We say that a sequence of distinct elements of is a cycle of if for any with (indices are modulo ). The length of the cycle is the number . In order to prove the claim, we show by a case analysis that has no cycles. Suppose that has a cycle of length 2. If and , then there is an element with and . We have from and that , a contradiction. If , then there exist two elements with and . Thus has a forbidden configuration for , a contradiction. If and , then there exist two elements with and . Thus has a forbidden configuration for , a contradiction. Therefore, . A similar argument would show that ; hence has no cycles of length 2. Suppose that has a cycle of length grater than 2. Let be such a cycle with minimal length, that is, there is no relation with . If there is an index with , then , a contradiction. Suppose that there is an index with and . Then there is an element with and . We have and , which imply , a contradiction. Suppose that there is an index with and . Then there is an element with and . Since in , the elements induce an order . Since fulfills the rule and , we have . We now have , a contradiction. Therefore, there is no index with . Suppose that there is an index with . Then there exist two elements with and . If then implies , a contradiction. If then the elements induce a forbidden configuration for , a contradiction. Thus , and we have from . If then the elements induce a forbidden configuration for , a contradiction. If then implies , a contradiction. Therefore, there is no index with . A similar argument would show that there is no index with . Suppose that there is an index with ; hence and . Then there exist three elements with and . We have ; for otherwise the elements induce a forbidden configuration for , a contradiction. If then the elements induce a forbidden configuration for , a contradiction. If then implies , a contradiction. Thus , and we have from . If then implies , a contradiction. If then the elements induce a forbidden configuration for , a contradiction. Therefore, there is no index with . A similar argument would show that there is no index with . Therefore, we have that the relation has no cycles, and thus the claim holds. Assume that the elements of are indexed so that if in . We define a function recursively as follows. For the base case, we set ; for an index with , we set

The function satisfies the following properties:

1. [label=()]

2. .

3. If then .

4. If then .

5. If then .

Trivially, the function satisfies the properties (a)–(c). We use the followings to show that satisfies the property (d). There is no three indices , , with such that and ; for otherwise we would have , which implies in , a contradiction. There is no three indices , , with such that and ; for otherwise we would have , which implies in , a contradiction. There is no four indices , , , with such that and ; for otherwise we would have , which implies , a contradiction. Suppose that there exist two indices and with and . If , we have , and three indices , , and satisfy , , and , a contradiction. If , we have from the definition of that there is an index with such that and . Then, , , and , a contradiction. Therefore, the function satisfies the property (d). Let be the set of intervals on the real line; the property (a) of the function ensures that each interval of exists. We have from the property (b) that no interval of properly contains another. Let be the semiorder obtained from the proper interval representation . The properties (c) and (d) imply that in if and in if . Therefore, . The relation