Linear-semiorders and their incomparability graphs
aa r X i v : . [ c s . D M ] J u l LINEAR-SEMIORDERS AND THEIR INCOMPARABILITY GRAPHS
ASAHI TAKAOKAA bstract . A linear-interval order is the intersection of a linear order and an interval order. For this class of orders, severalstructural results have been shown. In this paper, we study a natural subclass of linear-interval orders. We call a partial ordera linear-semiorder if it is the intersection of a linear order and a semiorder. We show a characterization of linear-semiordersin terms of linear extensions. This gives a vertex ordering characterization of their incomparability graphs. We also show thatbeing a linear-semiorder is a comparability invariant.
1. I ntroduction
A graph is an intersection graph if there is a set of objects such that each vertex corresponds to an object andtwo vertices are adjacent if and only if the corresponding objects have a nonempty intersection. Intersection graphs ofgeometric objects have been widely investigated due to their interesting structures and their applications. See [4,10,17]for survey.Well-known examples of intersection graphs are interval graphs and permutation graphs. An interval graph isthe intersection graph of intervals on the real line. Let L and L be two horizontal lines in the x y -plane with L above L . A permutation graph is the intersection graph of line segments joining a point on L and a point on L . Acommon generalization of the two graph classes is trapezoid graphs [6, 7]. An interval on L and an interval on L define a trapezoid between L and L . A trapezoid graph is the intersection graph of such trapezoids. The structure oftrapezoid graphs are well investigated, and many recognition algorithms are provided. See [12, 15, 17].There is a correspondence between partial orders and the intersection graphs of geometric objects between the twolines [11], [12, Theorem 1.11]. A partial order P on a set V is a trapezoid order if for each element v ∈ V , there isa trapezoid T ( v ) between L and L so that for any two elements u , v ∈ V , we have u ≺ v in P if and only if T ( u )lies completely to the left of T ( v ). The set of trapezoids { T ( v ) : v ∈ V } is called a trapezoid representation of P . Byrestricting the trapezoids in the representation, many classes of orders have been introduced [1, 2, 16].An up-triangle order [1] is a partial order representable by triangles spanned by a point on L and an interval on L . An up-triangle order is also known as a PI order [4–6], where PI stands for Point-Interval , and as a linear-intervalorder [14] since it is the intersection of a linear order and an interval order. We use in this paper the term linear-intervalorders to denote such orders. Several structural results have been shown for this class of orders [5, 6, 19], includingpolynomial-time recognition algorithms [14, 18]. As noted in [14], this is one of the first results on the recognition oforders that are the intersection of orders from two di ff erent classes.In this paper, we study up-triangle orders representable by triangles spanned by a point on L and a unit-length interval on L . See Fig. 1 for example. Such an order is the intersection of a linear order and a semiorder; hence wecall it a linear-semiorder . 2. P reliminaries A partially ordered set is a pair ( V , P ), where V is a set and P is a binary relation on V that is irreflexive, transitive,and therefore asymmetric. The set V is called the ground set while the relation P is called a partial order on V . In thispaper, we will deal only with partial orders on finite sets.We denote partial orders by ≺ instead of P , that is, we write u ≺ v in P if and only if ( u , v ) ∈ P . Two elements u , v ∈ V are comparable in P if u ≺ v or u ≻ v ; otherwise u and v are incomparable , which we denote u k v . A partialorder P on a set V is a linear order if any two elements of V are comparable in P . Date : July 19, 2019.2010
Mathematics Subject Classification.
Key words and phrases.
Comparability invariant, Linear-interval orders, PI graphs, Semiorders, Triangle orders, Vertex ordering characteriza-tion . b c ed f ( a ) a b c d e f L L ( b ) F igure
1. A partial order (the dual of chevron) with a triangle representation.A partial order P on a set V is an interval order if for each element v ∈ V , there is a (closed) interval I ( v ) on thereal line so that for any two elements u , v ∈ V , we have u ≺ v in P if and only if I ( u ) lies completely to the left of I ( v ). Here, the interval I ( u ) = [ l ( u ) , r ( u )] lies completely to the left of I ( v ) = [ l ( v ) , r ( v )], and we write I ( u ) ≪ I ( v ), if r ( u ) < l ( v ). The set of intervals { I ( v ) : v ∈ V } is called an interval representation of P .An interval representation is unit if every interval has unit length, and it is proper if no interval properly containsanother. An interval order is a semiorder if it has a unit interval representation. It is known that a partial order is asemiorder if and only if it has a proper interval representation [3].Let P and P be two partial orders on the same ground set V . The intersection of P and P is the partial order P = P ∩ P . Equivalently, the intersection of P and P is the partial order P on V such that u ≺ v in P if and only if u ≺ v in both P and P . We call an order a linear-semiorder if it is the intersection of a linear order and a semiorder.Let P be a partial order on a set V . The comparability graph of P is the graph G = ( V , E ) such that u v ∈ E if andonly if u and v are comparable in P ; the incomparability graph of P is the graph G = ( V , E ) such that u v ∈ E if and onlyif u k v in P . A cocomparability graph is the complement of a comparability graph. Note that any cocomparabilitygraph is the incomparability graph of some partial order.3. C omparability invariance A property of partial orders is a comparability invariant if either all orders with the same comparability graph havethat property or none have that property. It is known that being a linear-interval order is a comparability invariant [5].In this section, we will show the following.
Theorem 1.
Being a linear-semiorder is a comparability invariant.
We use the proof technique developed in [8].Let P be a partial order on a set V . A non-empty subset A ⊂ V is autonomous in P if for any element v ∈ V − A ,whenever v ≺ a , v k a , or v ≻ a holds in P for some element a ∈ A , then the same holds for all elements a ∈ A . Let P ′ be a partial order having the same comparability graph as P . The order P ′ is obtained by a reversal from P if there isan autonomous set A of P such that:(1) If not both of u and v are in A , then u ≺ v in P ′ if and only if u ≺ v in P .(2) If both of u and v are in A , then u ≺ v in P ′ if and only if u ≻ v in P .We denote by P | A the order obtained from P by reversing A .The following theorem [9] provides a simple scheme to show the comparability invariance results: Two orders P and P ′ have the same comparability graph if and only if there is a finite sequence of orders P , P , . . . , P k such that P = P , P k = P ′ , and P i is obtained from P i − by a reversal for each i = , , . . . , k . Therefore, in order to proveTheorem 1, we will show the following claim. Claim. If an order P on a set V is a linear-semiorder and a subset A ⊂ V is autonomous in P, then P | A is alinear-semiorder.
Recall that L and L are two horizontal lines in the x y -plane with L above L . As a representation of a linear-semiorder, we use a set of triangles between L and L as follows. et P be a linear-semiorder on a set V , and let L and S be a linear order and a semiorder on V with L ∩ S = P . Let { p ( v ) : v ∈ V } be a set of points on L such that p ( u ) < p ( v ) if and only if u ≺ v in L for any two elements u , v ∈ V .Notice that all the points are distinct by definition. Let { I ( v ) : v ∈ V } be a unit interval representation of S on L . Weassume that no two intervals share a common endpoint. Let I ( v ) = [ l ( v ) , r ( v )] for each element v ∈ V . Since eachinterval has unit length, l ( v ) + = r ( v ).Let T ( v ) be the triangle spanned by p ( v ) and I ( v ). A triangle T ( u ) lies completely to the left of T ( v ), and we write T ( u ) ≪ T ( v ), if p ( u ) < p ( v ) on L and I ( u ) ≪ I ( v ) on L . We have that u ≺ v in P if and only if T ( u ) ≪ T ( v ) for anytwo elements u , v ∈ V ; hence we call the set { T ( v ) : v ∈ V } a triangle representation of P .Note that in the following, we use the term triangle to denote a triangle spanned by a point on L and a unit-lengthinterval on L .Now, we start to prove Theorem 1. We fix a pair of a linear-semiorder P and an autonomous set A of P . We also fixa triangle representation { T ( v ) : v ∈ V } of P .An element a ∈ A is isolated if a k a ′ in P for any element a ′ ∈ A − { a } . Let A ∗ be the subset of A obtained byremoving all isolated elements of A . We can observe the following. Lemma 2.
The set A ∗ is autonomous in P, and P | A ∗ = P | A. Thus we assume without loss of generality A ∗ , ∅ .We define C ( A ∗ ) as the convex region spanned by the triangles T ( v ) with v ∈ A ∗ . We also define that l = min v ∈ A ∗ p ( v ) and r = max v ∈ A ∗ p ( v ), similarly, l = min v ∈ A ∗ l ( v ) and r = max v ∈ A ∗ r ( v ). Notice that C ( A ∗ ) is the trapezoidspanned by the interval [ l , r ] on L and the interval [ l , r ] on L , while l is the upper left corner of C ( A ∗ ) and r isthe upper right corner of C ( A ∗ ), similarly, l is the lower left corner of C ( A ∗ ) and r is the lower right corner of C ( A ∗ ).Moreover, let a l and a r denote the elements of A ∗ with p ( a l ) = l and p ( a r ) = r , respectively. Similarly, let a l and a r denote the elements of A ∗ with l ( a l ) = l and r ( a r ) = r , respectively.Let B be the set of elements v ∈ V with T ( v ) ⊂ C ( A ∗ ). Obviously, B ⊃ A ∗ . The convex region C ( B ) spanned by thetriangles T ( v ) with v ∈ B is equal to C ( A ∗ ). Lemma 3.
Let v ∈ V − B. If T ( v ) ∩ C ( A ∗ ) , ∅ then T ( v ) intersects with every triangle contained in C ( A ∗ ) . Note that the triangles in the lemma are not just the triangles of the representation { T ( v ) : v ∈ V } of P , but everytriangle spanned by a point p on L with p ∈ [ l , r ] and a unit-length interval I on L with I ⊂ [ l , r ]. Proof.
Since v < B , we have T ( v ) C ( A ∗ ), and hence p ( v ) < [ l , r ] or I ( v ) [ l , r ].Suppose p ( v ) < l . Since T ( v ) ∩ C ( A ∗ ) , ∅ , we have l < r ( v ). Thus T ( v ) ∩ T ( a l ) , ∅ , and hence v k a l in P .Since A ∗ is autonomous, v k a r in P , and hence T ( v ) ∩ T ( a r ) , ∅ . Thus l ( a r ) < r ( v ). Therefore, we have p ( v ) < l and l ( a r ) = r − < r ( v ), and the claim holds. A similar argument would show that the claim holds when r < p ( v ).Suppose l ( v ) < l . If l < r ( v ) then T ( v ) ∩ T ( a l ) , ∅ . If r ( v ) < l then T ( v ) ∩ C ( A ∗ ) , ∅ implies l < p ( v ), and hence T ( v ) ∩ T ( a l ) , ∅ . Thus in both cases, v k a in P for some element a ∈ A ∗ . Since A ∗ is autonomous, v k a r in P . Since a r is not isolated, there is an element a ′ ∈ A ∗ with T ( a ′ ) ≪ T ( a r ). Since l ( v ) < l < l ( a ′ ) and each interval has unitlength, r ( v ) < r ( a ′ ), and hence I ( v ) ≪ I ( a r ). It follows that p ( a r ) < p ( v ). Therefore, we have r < p ( v ) and l ( v ) < l ,and the claim holds. A similar argument would show that the claim holds when r < r ( v ). (cid:3) As a consequence of Lemma 3, we have the key lemma.
Lemma 4.
For any element v ∈ V − B, one of the following holds: – T ( v ) lies completely to the left of C ( A ∗ ) . – T ( v ) intersects with every triangle contained in C ( A ∗ ) . – T ( v ) lies completely to the right of C ( A ∗ ) . We also have the following from Lemma 4
Lemma 5.
The set B is autonomous in P.Proof.
Let v ∈ V − B . Suppose v k b in P for some element b ∈ B . Then T ( v ) ∩ T ( b ) , ∅ implies T ( v ) ∩ C ( A ∗ ) , ∅ , andhence T ( v ) ∩ T ( b ) , ∅ for all elements b ∈ B . Thus v k b in P for all elements b ∈ B . If v ≺ b in P for some element b ∈ B , then T ( v ) lies completely to the left of C ( A ∗ ), and hence v ≺ b in P for all elements b ∈ B . If v ≻ b in P forsome element b ∈ B , then T ( v ) lies completely to the right of C ( A ∗ ), and hence v ≻ b in P for all elements b ∈ B . (cid:3) et P = P | B . We define { T ′ ( v ) : v ∈ V } as the set of triangles obtained from the representation { T ( v ) : v ∈ V } of P by flipping the representation of B relative to C ( B ), that is, T ′ ( v ) = T ( v ) for any element v ∈ V − B while for eachelement v ∈ B , the triangle T ′ ( v ) is the triangle spanned by the point l + r − p ( v ) and the interval [ l + r − r ( v ) , l + r − l ( v )]. The convex region C ′ ( B ) spanned by the triangles T ′ ( v ) with v ∈ B is equal to C ( B ). Lemma 6.
The order P is a linear-semiorder with a representation { T ′ ( v ) : v ∈ V } .Proof. Let P ′ be the partial order on V defined by the representation { T ′ ( v ) : v ∈ V } . If neither u nor v is in B , thentheir relation in P ′ remains the same as in P . If one of u and v is in B , then Lemma 4 ensures that their relation in P ′ remains the same as in P . If both u and v are in B , then T ′ ( u ) ≪ T ′ ( v ) ⇐⇒ T ( u ) ≫ T ( v ); hence u ≺ v in P ′ if andonly if u ≻ v in P . Thus P ′ = P . (cid:3) Let A = B − A ∗ . If A = ∅ then B = A ∗ . It follows that P | A = P | A ∗ = P | B = P is a linear-semiorder. Thus weassume without loss of generality A , ∅ . Lemma 7.
All the elements of A is incomparable to all the elements of A ∗ in P.Proof. Suppose that there is an element a ∈ A with a ≺ a ′ in P for some element a ′ ∈ A ∗ . Since A ∗ is autonomous, a ≺ a l and a ≺ a l in P , contradicting to T ( a ) ∩ C ( A ∗ ) , ∅ . Similarly, we have that there is no elements a ∈ A with a ≻ a ′ in P for some element a ′ ∈ A ∗ . (cid:3) Lemma 8.
The set A is autonomous in P , and P | A = P | A.Proof.
Let v ∈ V − A . If v ∈ V − B then, since B is autonomous in P , whenever v ≺ a , v k a , or v ≻ a holds in P forsome element a ∈ A ⊂ B , then the same holds for all elements a ∈ A . If v ∈ B − A then v ∈ A ∗ , and we have fromLemma 7 that v k a in P for all elements a ∈ A . Thus A is autonomous in P .Lemma 7 implies P | B = P | A ∗ | A . Thus P | B | A = P | A ∗ . Since P = P | B and P | A ∗ = P | A , we have P | A = P | A . (cid:3) We repeat the process replacing P and A with P and A . Let A ∗ be the subset of A obtained by removing allisolated elements of A . By Lemma 2, the set A ∗ is autonomous in P and P | A ∗ = P | A . We assume without lossof generality A ∗ , ∅ . We define C ′ ( A ∗ ) as the convex region spanned by the triangles T ′ ( v ) with v ∈ A ∗ . Let B be theset of elements v ∈ V with T ′ ( v ) ⊂ C ′ ( A ∗ ). By Lemma 5, the set B is autonomous in P , and let P = P | B . ByLemma 6, the order P is a linear-semiorder. Let A = B − A ∗ , and we assume without loss of generality A , ∅ . ByLemma 8, the set A is autonomous in P , and P | A = P | A .Let A ∗ be the subset of A obtained by removing all isolated elements of A . By Lemma 2, the set A ∗ is autonomousin P , and P | A ∗ = P | A . Lemma 9.
The set A ∗ is a proper subset of A ∗ .Proof. Since A ∗ ⊂ B , we have C ′ ( A ∗ ) ⊂ C ′ ( B ) = C ( B ) = C ( A ∗ ), and hence B ⊂ B . We also have that the inclusion isproper since, for example, a l < B .Recall that B can be partitioned into three sets A ∗ , A − A ∗ , and A ∗ . By definition, all the elements A − A ∗ areisolated in A . By Lemma 7, all the elements of A − A ∗ are incomparable to all the elements of A ∗ . Thus they areisolated in B .We have A = B − A ∗ ⊂ B − A ∗ = A ∗ ∪ ( A − A ∗ ). Since all the elements of A − A ∗ are isolated in B ⊃ A , wehave A ∗ ∩ ( A − A ∗ ) = ∅ and A ∗ ⊂ A ∗ . Since a l ∈ A ∗ but a l < B ⊃ A ∗ , the inclusion is proper. (cid:3) Now, we are ready to prove the main theorem.
Proof of Theorem 1.
Suppose that the theorem does not hold. It follows that there is a pair of a linear-semiorder P andan autonomous set A of P such that P | A is not a linear-semiorder. Among such pairs, we choose one such that | A | isminimal.We have from Lemma 8 that P | A = P | A = P | A = P | A ∗ , but we also have from Lemma 9 that | A ∗ | < | A ∗ | ≤ | A | , contradicting to the minimality of A . (cid:3) y z w ( a ) x y z w ( b ) x y z ( c ) F igure
2. (a) A forbidden configuration for + . (b) A forbidden configuration for + . (c) Apartial order with ( x , y ) , ( y, z ) ∈ R ∪ R but ( x , z ) < R ∪ R . An arrow u → v denotes that u ≺ v in P ; a dashed arrow u d v denotes that u ≺ v in L but u k v in P .4. C haracterization Let P be a partial order on a set V . A linear order L on V is a linear extension of P if u ≺ v in L whenever u ≺ v in P . Hence, the linear extension L of P has all the relations of P with the additional relations that make L linear. Wedefine some properties of linear extensions.The order + of P is the partial order consisting of four elements x , y , z , w of V such that x ≺ y and z ≺ w while x k w and z k y in P . Notice that x k z and y k w in P ; for otherwise we would have x ≺ w or z ≺ y in P . We say that alinear extension L of P fulfills the + rule if y ≺ z or w ≺ x in L for each induced suborder + of P ; if y ≺ z in L then x ≺ y ≺ z ≺ w in L ; if w ≺ x in L then z ≺ w ≺ x ≺ y in L . Equivalently, a linear extension L of P is said to fulfillthe + rule if there is no four elements x , y, z , w of V such that x ≺ y , z ≺ w , x k w , and z k y in P while x ≺ w and z ≺ y in L . See Fig. 2(a). We call such an induced suborder a forbidden configuration for + .The order + of P is the partial order consisting of four elements x , y , z , w of V such that x ≺ y ≺ z while x k w and w k z in P . Notice that y k w in P ; for otherwise we would have x ≺ w or w ≺ z in P . We say that a linearextension L of P fulfills the + rule if w ≺ x or z ≺ w in L for each induced suborder + of P ; if w ≺ x in L then w ≺ x ≺ y ≺ z in L ; if z ≺ w in L then x ≺ y ≺ z ≺ w in L . Equivalently, a linear extension L of P is said to fulfill the + rule if there is no four elements x , y, z , w of V such that x ≺ y ≺ z , x k w , and w k z in P while x ≺ w ≺ z in L .See Fig. 2(b). We call such an induced suborder a forbidden configuration for + .Our previous work [19] shows that a partial order is a linear-interval order if and only if it has a linear extensionfulfilling the + rule. In this paper, we show a similar characterization for linear-semiorders. Theorem 10.
A partial order is a linear-semiorder if and only if it has a linear extension fulfilling the + rule and + rule.Proof. The necessity and su ffi ciency follow immediately from Lemmas 11 and 12, respectively. (cid:3) Lemma 11.
If a partial order P on a set V has a linear order L and a semiorder S with L ∩ S = P, then L has noforbidden configurations for + nor + .Proof. It is shown in [6, 19] that L has no forbidden configurations for + . Suppose for a contradiction that L hasa forbidden configuration for + consisting of four elements x , y, z , w of V such that x ≺ y ≺ z , x k w , and z k w in P while x ≺ w ≺ z in L . Let { I ( v ) : v ∈ V } be a proper interval representation of S , and let I ( v ) = [ l ( v ) , r ( v )] for eachelement v of V . Since x ≺ y ≺ z in P , we have r ( x ) < l ( y ) ≤ r ( y ) < l ( z ). Since x ≺ w in L and x k w in P , we have x ⊀ w in S . Thus l ( w ) ≤ r ( x ). Similarly, since w ≺ z in L and w k z in P , we have w ⊀ z in S . Thus l ( z ) ≤ r ( w ). Therefore, wehave I ( w ) ⊃ I ( y ), contradicting to that no interval properly contains another. (cid:3) Lemma 12.
If a partial order P on a set V has a linear extension L fulfilling the + rule and + rule, then thereis a semiorder S with L ∩ S = P. The semiorder S and its proper interval representation can be obtained in O ( n ) time, where n is the number of elements of V. roof. 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 ( u , v ) ∈ P if u ≺ v in P ; we write ( u , v ) ∈ L − P if u k v in P and u ≺ v in L .Suppose that there is such a semiorder S . Let { I ( v ) : v ∈ V } be a proper interval representation of S , and let I ( v ) = [ l ( v ) , r ( v )] for each element v of V . Let x and y be two elements of V . Suppose that there is an element z ∈ V with ( x , z ) ∈ P and ( y, z ) ∈ L − P . Then ( y, z ) < S and hence l ( z ) ≤ r ( y ). Since ( x , z ) ∈ P implies r ( x ) < l ( z ), wehave r ( x ) < r ( y ). Similarly, if there is an element z ∈ V with ( z , x ) ∈ L − P and ( z , y ) ∈ P , then l ( x ) < l ( y ), and hence r ( x ) < r ( y ). Trivially, r ( x ) < r ( y ) if ( x , y ) ∈ P . To capture these relations, we define the following.Let R be the binary relation on V such that ( x , y ) ∈ R if there is an element z ∈ V with ( x , z ) ∈ P and ( y, z ) ∈ L − P ;let R be the binary relation on V such that ( x , y ) ∈ R if there is an element z ∈ V with ( z , x ) ∈ L − P and ( z , y ) ∈ P .Let Q = P ∪ R ∪ R . Note that the binary relation Q is not transitive in general. See for example the partial order inFig. 2(c); we have ( x , y ) ∈ R and ( y, z ) ∈ R , but ( x , z ) < R ∪ R . Claim. There is a linear order L Q on V such that x ≺ y in L Q whenever ( x , y ) ∈ Q. We say that a sequence of distinct elements v , v , . . . , v k − of V is a cycle of Q if ( v i , v i + ) ∈ Q for any i with0 ≤ i < k (indices are modulo k ). The length of the cycle is the number k . In order to prove the claim, we show by acase analysis that Q has no cycles.Suppose that Q has a cycle of length 2. If ( v , v ) ∈ R and ( v , v ) ∈ P , then there is an element u ∈ V with( v , u ) ∈ P and ( v , u ) ∈ L − P . We have from ( v , v ) ∈ P and ( v , u ) ∈ P that ( v , u ) ∈ P , a contradiction. If( v , v ) , ( v , v ) ∈ R , then there exist two elements u , u ∈ V with ( v , u ) , ( v , u ) ∈ P and ( v , u ) , ( v , u ) ∈ L − S .Thus L has a forbidden configuration for + , a contradiction. If ( v , v ) ∈ R and ( v , v ) ∈ R , then there exist twoelements u , u ∈ V with ( v , u ) , ( u , v ) ∈ P and ( v , u ) , ( u , v ) ∈ L − S . Thus L has a forbidden configuration for + , a contradiction. Therefore, ( v , v ) < R . A similar argument would show that ( v , v ) < R ; hence Q has nocycles of length 2.Suppose that Q has a cycle of length grater than 2. Let v , v , . . . , v k − be such a cycle with minimal length, that is,there is no relation ( v i , v j ) ∈ Q with j , i +
1. If there is an index i with ( v i , v i + ) , ( v i + , v i + ) ∈ P , then ( v i , v i + ) ∈ P ,a contradiction. Suppose that there is an index i with ( v i , v i + ) ∈ P and ( v i + , v i + ) ∈ R . Then there is an element u ∈ V with ( v i + , u ) ∈ P and ( v i + , u ) ∈ L − P . We have ( v i , u ) ∈ P and ( v i + , u ) ∈ L − P , which imply ( v i , v i + ) ∈ R , acontradiction. Suppose that there is an index i with ( v i , v i + ) ∈ P and ( v i + , v i + ) ∈ R . Then there is an element u ∈ V with ( u , v i + ) ∈ L − P and ( u , v i + ) ∈ P . Since v i k v i + in P , the elements u , v i + , v i , v i + induce an order + . Since L fulfills the + rule and ( u , v i + ) ∈ L − P , we have ( u , v i ) ∈ L − P . We now have ( v i , v i + ) ∈ R , a contradiction.Therefore, there is no index i with ( v i , v i + ) ∈ P .Suppose that there is an index i with ( v i , v i + ) , ( v i + , v i + ) ∈ R . Then there exist two elements u , u ∈ V with( v i , u ) , ( v i + , u ) ∈ P and ( v i + , u ) , ( v i + , u ) ∈ L − S . If ( v i , u ) ∈ P then ( v i + , u ) ∈ L − P implies ( v i , v i + ) ∈ R ,a contradiction. If ( v i , u ) ∈ L − P then the elements v i , u , v i + , u induce a forbidden configuration for + , acontradiction. Thus ( u , v i ) ∈ L , and we have ( v i + , u ) ∈ L from ( v i + , u ) , ( v i , u ) ∈ L . If ( v i + , u ) ∈ P then the elements v i + , u , v i + , u induce a forbidden configuration for + , a contradiction. If ( v i + , u ) ∈ L − P then ( v i , u ) ∈ P implies( v i , v i + ) ∈ R , a contradiction. Therefore, there is no index i with ( v i , v i + ) , ( v i + , v i + ) ∈ R . A similar argument wouldshow that there is no index i with ( v i , v i + ) , ( v i + , v i + ) ∈ R .Suppose that there is an index i with ( v i , v i + ) ∈ R ; hence ( v i + , v i + ) ∈ R and ( v i + , v i + ) ∈ R . Then there existthree elements u , u , u ∈ V with ( v i , u ) , ( u , v i + ) , ( v i + , u ) ∈ P and ( v i + , u ) , ( u , v i + ) , ( v i + , u ) ∈ L − P . We have u , u ; for otherwise the elements u , v i + , u = u , v i + induce a forbidden configuration for + , a contradiction.If ( v i + , u ) ∈ P then the elements u , v i + , u , v i + induce a forbidden configuration for + , a contradiction. If( v i + , u ) ∈ L − P then ( v i , u ) ∈ P implies ( v i , v i + ) ∈ R , a contradiction. Thus ( u , v i + ) ∈ L , and we have ( v i + , u ) ∈ L from ( v i + , u ) , ( v i + , u ) ∈ L . If ( v i + , u ) ∈ P then ( v i + , u ) ∈ L − P implies ( v i + , v i + ) ∈ R , a contradiction.If ( v i + , u ) ∈ L − P then the elements u , v i + , u , v i + induce a forbidden configuration for + , a contradiction.Therefore, there is no index i with ( v i , v i + ) ∈ R . A similar argument would show that there is no index i with( v i , v i + ) ∈ R .Therefore, we have that the relation Q has no cycles, and thus the claim holds.Assume that the elements v , v , . . . , v n of V are indexed so that i < j if v i ≺ v j in L Q . We define a function f : { , , . . . , n } → N recursively as follows. For the base case, we set f (1) =
0; for an index i with 1 < i ≤ n , we set f ( i ) = max { j : ( v j , v i ) ∈ P } if there is an index j > f ( i −
1) with ( v j , v i ) ∈ P , f ( i −
1) otherwise. c ef bd ( a ) ac e f b d ( b ) F igure
3. (a) The semiorder S . (b) The proper interval representation of S . Claim. The function f satisfies the following properties: (a) 0 ≤ f ( i ) < i. (b) If i < j then f ( i ) ≤ f ( j ) . (c) If ( v i , v j ) ∈ P then i ≤ f ( j ) . (d) If ( v i , v j ) ∈ L − P then f ( j ) < i. Trivially, the function f satisfies the properties (a)–(c). We use the followings to show that f satisfies the property(d).There is no three indices i , j , k with i < j < k such that ( v i , v k ) ∈ L − P and ( v j , v k ) ∈ P ; for otherwise we wouldhave ( v j , v i ) ∈ R , which implies v j ≺ v i in L Q , a contradiction. There is no three indices i , j , k with i < j < k such that( v i , v k ) ∈ L − P and ( v i , v j ) ∈ P ; for otherwise we would have ( v k , v j ) ∈ R , which implies v k ≺ v j in L Q , a contradiction.There is no four indices i , j , k , h with i < j < k < h such that ( v i , v h ) ∈ L − P and ( v j , v k ) ∈ P ; for otherwise we wouldhave ( v h , v j ) , ( v k , v i ) ∈ L , which implies ( v h , v i ) ∈ L , a contradiction.Suppose that there exist two indices i and j with ( v i , v j ) ∈ L − P and i ≤ f ( j ). If ( v f ( j ) , v j ) ∈ P , we have i , f ( j ),and three indices i , f ( j ), and j satisfy i < f ( j ) < j , ( v i , v j ) ∈ L − P , and ( v f ( j ) , v j ) ∈ P , a contradiction. If ( v f ( j ) , v j ) < P ,we have from the definition of f that there is an index k with k < j such that f ( k ) = f ( j ) and ( v f ( k ) , v k ) ∈ P . Then, i ≤ f ( k ) < k < j , ( v i , v j ) ∈ L − P , and ( v f ( k ) , v k ) ∈ P , a contradiction. Therefore, the function f satisfies the property(d).Let I = { [ f ( i ) + i − n , i ] : 1 ≤ i ≤ n } be the set of intervals on the real line; the property (a) of the function f ensuresthat each interval of I exists. We have from the property (b) that no interval of I properly contains another. Let S bethe semiorder obtained from the proper interval representation I . The properties (c) and (d) imply that v i ≺ v j in S if( v i , v j ) ∈ P and v i ⊀ v j in S if ( v i , v j ) ∈ L − P . Therefore, L ∩ S = P .The relation Q is obtained in O ( n ) time from P and L . The function f and the representation I of S is obtained in O ( n ) time from L Q . (cid:3) Using an example, we illustrate the construction of the semiorder in the proof of Lemma 12. Consider the linear-semiorder P in Fig. 1. Let L be the linear order such that a ≺ b ≺ c ≺ d ≺ e ≺ f in L ; we can observe that L fulfillsthe + rule and + rule. We have that R ∪ R = { ( a , c ) , ( c , b ) , ( c , e ) , ( e , b ) , ( e , d ) , ( f , d ) } . Thus there is two linearorders that contains all the ordered pairs of Q = P ∪ R ∪ R , that is, the linear order with a ≺ c ≺ e ≺ b ≺ f ≺ d andthe linear order with a ≺ c ≺ e ≺ f ≺ b ≺ d . We choose the latter as L Q . We now obtain the intervals I ( a ) = [0 , I ( c ) = [ , I ( e ) = [1 + , I ( f ) = [3 + , I ( b ) = [3 + , I ( d ) = [5 + , S defined by the intervals is that shown in Fig. 3(a), and wecan observe L ∩ S = P .We finally show two byproducts of the characterization, one of which is a vertex ordering characterization of theincomparability graphs of linear-semiorders, and the other is related to the hierarchy of classes of orders.Let G = ( V , E ) be a graph. A vertex ordering of G is a linear order of the vertex set V . A vertex orderingcharacterization of a graph class is a characterization of the following type: a graph G is in that class if and only if G has a vertex ordering fulfilling some properties. For example, a graph G is a cocomparability graph if and only if thereis a vertex ordering L of G such that for any three vertices u , v, w of G with u ≺ v ≺ w in L , if u w ∈ E then u v ∈ E or a ) ( b ) ( c )( d ) ( e ) F igure
4. Forbidden patterns. Lines and dashed lines denote edges and non-edges, respectively.Edges that may or may not be present is not drawn. vw ∈ E [13]. Equivalently, a graph is a cocomparability graph if and only if it has a vertex ordering that contains nosuborderings in Fig. 4(a).Incomparability graphs of linear-interval orders can be characterized so that a graph G is such a graph if and onlyif there is a vertex ordering of G that contains no suborderings in Figs. 4(a)–(c) [19]. For linear-semiorders, a similarcharacterization follows from Theorem 10. Corollary 13.
A graph G is the incomparability graph of a linear-semiorder if and only if there is a vertex orderingof G that contains no suborderings in Figs. 4(a)–(e).Proof.
Assume that there is a linear-semiorder P on a set V such that the incomparability graph of P is G . Then P has a linear extension L fulfilling the + rule and + rule. Notice that L can be regarded as a vertex ordering of G . Since L is a linear extension of a partial order, it has no suborderings in Fig. 4(a) as a vertex ordering of G . Since L has no forbidden configurations for + , it has no suborderings in Figs. 4(b) and 4(c). Since L has no forbiddenconfigurations for + , it has no suborderings in Figs. 4(d) and 4(e).Conversely, assume that there is a vertex ordering L of G = ( V , E ) that has no suborderings in Figs. 4(a)–(e). Let P be a binary relation on V such that ( u , v ) ∈ P if and only if for any two vertices u and v of G , we have u v < E and u ≺ v in L . Since L has no suborderings in Fig. 4(a), the relation P is transitive, and hence a partial order. Now,notice that L can be regarded as a linear extension of P . Since L has no suborderings in Figs. 4(b) and 4(c), it has noforbidden configurations for + as a linear extension of P . Since L has no suborderings in Figs. 4(d) and 4(e), it hasno forbidden configurations for + . (cid:3) The class of linear-interval orders contains interval orders and orders of dimension 2 as proper subclasses [6]. Thefollowing example shows that the class of interval orders is not a subclass of linear-semiorders.
Example 14.
The interval order P I in Fig. 5(a) is not a linear-semiorder.Proof. The interval representation of P I is shown in Fig. 5(b). Suppose that P I has a linear extension L fulfilling the + rule.The order + consisting of b , b , b , a requires that a ≺ b in L if and only if a ≺ b in L . The order + consisting of a , a , a , b requires that a ≺ b in L if and only if a ≺ b in L . The order + consisting of b , b , b , a requires that a ≺ b in L if and only if a ≺ b in L . Thus a ≺ b in L if and only if a ≺ b in L .The order + consisting of b , b , c , a requires that a ≺ b in L if and only if a ≺ c in L . The order + consisting of a , b , b , c requires that a ≺ c in L if and only if b ≺ c in L . The order + consisting of c , b , b , c requires that b ≺ c in L if and only if c ≺ c in L . Thus a ≺ b in L if and only if c ≺ c in L .Similarly, the order + consisting of c , b , b , a requires that a ≺ b in L if and only if a ≺ c in L . The order + consisting of b , b , a , c requires that a ≺ c in L if and only if b ≺ c in L . The order + consisting of b , b , c , c requires that b ≺ c in L if and only if c ≺ c in L . Thus a ≺ b in L if and only if c ≺ c in L , acontradiction. Therefore, P I has no linear extension fulfilling the + rule. (cid:3) Note that one can check by inspection that the interval order P I in Fig. 5(a) is a minimal forbidden order, that is,any induced suborder of P I is a linear-semiorder. a a b b b b b c c ( a ) a a a b b b b b c c ( b ) F igure
5. (a) An interval order P I that is not a linear-semiorder. (b) An interval representation of P I . Corollary 15.
The classes of linear-semiorders and interval orders are incomparable; therefore, the class of linear-semiorders is a proper subclass of linear-interval orders. The class of linear-semiorders contains semiorders andorders of dimension 2 as proper subclasses.Proof.
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