Dimension of posets with planar cover graphs excluding two long incomparable chains
David M. Howard, Noah Streib, William T. Trotter, Bartosz Walczak, Ruidong Wang
DDIMENSION OF POSETS WITH PLANAR COVER GRAPHSEXCLUDING TWO LONG INCOMPARABLE CHAINS
DAVID M. HOWARD, NOAH STREIB, WILLIAM T. TROTTER, BARTOSZ WALCZAK,AND RUIDONG WANG
Abstract.
It has been known for more than 40 years that there are posets with planar covergraphs and arbitrarily large dimension. Recently, Streib and Trotter proved that such posetsmust have large height. In fact, all known constructions of such posets have two large disjointchains with all points in one chain incomparable with all points in the other. Gutowski andKrawczyk conjectured that this feature is necessary. More formally, they conjectured that forevery k (cid:62)
1, there is a constant d such that if P is a poset with a planar cover graph and P excludes k + k , then dim( P ) (cid:54) d . We settle their conjecture in the affirmative. We also discusspossibilities of generalizing the result by relaxing the condition that the cover graph is planar. Introduction
We assume that the reader is familiar with basic notation and terminology for posets, includingsubposets, chains and antichains, minimal and maximal elements, linear extensions, orderdiagrams, and cover graphs. Extensive background information on the combinatorics of posetscan be found in [25, 26]. We will also assume that the reader is familiar with basic concepts ofgraph theory, including subgraphs, induced subgraphs, paths and cycles, and planar graphs.A subposet Q of P is convex if y ∈ Q whenever x, z ∈ Q and x < y < z in P . When Q is aconvex subposet of P , the cover graph of Q is an induced subgraph of the cover graph of P .Traditionally, the elements of a poset are called points , and this is what we do in this paper.Dushnik and Miller [6] defined the dimension of a poset P , denoted by dim( P ), as the leastpositive integer d for which there is a family R = { L , . . . , L d } of linear extensions of P suchthat x (cid:54) y in P if and only if x (cid:54) y in all L , . . . , L d . Clearly, if Q is a subposet of P , thendim( Q ) (cid:54) dim( P ). A poset has dimension 1 if and only it is a chain.For d (cid:62)
2, the standard example S d is the poset of height 2 consisting of d minimal elements a , . . . , a d and d maximal elements b , . . . , b d with a i < b j in S d if and only if i = j . As noted in [6],dim( S d ) = d for every d (cid:62)
2. So every poset that contains a large standard example has largedimension. On the other hand, it is well known that there are posets that have large dimensionbut do not contain the standard example S (see the more comprehensive discussion in [1]).In recent years, there have been a series of research papers exploring connections betweenthe dimension of a poset P and graph-theoretic properties of the cover graph of P . This papercontinues with that theme. A poset P is planar if it has a drawing with no edge crossings in itsorder diagram. All planar posets have planar cover graphs, and it is well known that there arenon-planar posets with planar cover graphs (see [25], page 67).It is an easy exercise to show that the standard example S d is a planar poset when 2 (cid:54) d (cid:54) S d is non-planar when d (cid:62)
5. However, in [24], it is shown that for every d (cid:62)
5, the non-planar poset S d is a subposet of a poset with a planar cover graph. Subsequently, A journal version of this paper appeared in
J. Comb. Theory Ser. A a r X i v : . [ m a t h . C O ] D ec D. M. HOWARD, N. STREIB, W. T. TROTTER, B. WALCZAK, AND R. WANG a b a b a b a b a b Figure 1.
Kelly’s example of a planar poset containing the standard example S as a subposetKelly [18] proved the stronger result: for every d (cid:62)
5, the non-planar poset S d is a subposet of aplanar poset P with dim( P ) = d (see Figure 1).In this paper, we do not distinguish between isomorphic posets, and we say that P contains Q when there is a subposet of P that is isomorphic to Q . Also, we say P excludes Q when P does not contain Q . For a positive integer k , we let k denote a k -element chain, and we let k + k denote a poset consisting of two chains of size k with all points in one chain incomparable withall points in the other. The above-mentioned constructions of posets with planar cover graphsand arbitrarily large dimension raise the following questions. Question 1.1.
Which of the following statements are true for every poset P with a planar covergraph and sufficiently large dimension?(1) P has many minimal elements;(2) P has large height, that is, P contains k for some large value of k ;(3) P contains k + k for some large value of k ;(4) P contains S k for some large value of k .The construction in [24] shows that for every d (cid:62)
2, there is a poset with dimension d , aunique minimal element, a unique maximal element, and a planar cover graph. On the otherhand, in [29], the following result is proved for planar posets. Theorem 1.2. If P is a planar poset with t minimal elements, then dim( P ) (cid:54) t + 1 . Furthermore, it is shown in [29] that this inequality is tight when t = 1 and t = 2. However,when t (cid:62)
3, it is only known that there are planar posets with t minimal elements that havedimension t + 3. Since a poset and its dual have the same dimension, entirely analogousstatements can be made about maximal elements.The second question was answered in the affirmative in [23], where the following theorem(restated in a form consistent with the results of this paper) is proved. Theorem 1.3.
For every positive integer k , there is an integer d such that if P is a poset thatexcludes k and the cover graph of P is planar, then dim( P ) (cid:54) d . The bound on d from [23] is very weak, due to extensive use of Ramsey theory in the proof;however, greatly improved bounds are available via [21]. Furthermore, it is shown in [15] thatplanar posets excluding k have dimension bounded by O ( k ). IMENSION OF POSETS WITH PLANAR COVER GRAPHS EXCL. TWO LONG INCOMP. CHAINS 3
Gutowski and Krawczyk [11] posed the third question and conjectured that it should alsohave an affirmative answer. In this paper, we will settle their conjecture in the affirmative byproving the following theorem, which is the main result of this paper.
Theorem 1.4.
For every positive integer k , there is an integer d such that if P is a poset thatexcludes k + k and the cover graph of P is planar, then dim( P ) (cid:54) d . While the conjecture of Gutowski and Krawczyk might seem entirely natural just fromreflecting on the properties of the Kelly construction, it was also motivated by the results of[4, 5, 7, 17, 20], where combinatorial properties of posets excluding k + k played a central role.The fourth question, which was apparently first raised in [25] (see the comment on page 119),remains open, and we consider it one of the central challenges in this area of research. Mostresearchers feel that the answer is again “yes”. Formally, we can state the following conjecture. Conjecture 1.5.
For every positive integer k , there exists an integer d such that if P is a posetthat excludes the standard example S k and the cover graph of P is planar, then dim( P ) (cid:54) d . In the next section, we provide a brief summary of notation, terminology, and backgroundmaterial. This discussion applies to any research problem involving dimension. Then, in Section 3,we develop some properties of the class of posets that exclude k + k . As these results may findapplication to other combinatorial problems for posets, the results of that section are presentedfor posets in general—with no assumption that the cover graph is planar. The proof of ourmain theorem is given in the next three sections. Finally, in Section 7, we discuss possibilitiesof generalizing Theorem 1.4 and Conjecture 1.5 beyond planarity—to posets that have covergraphs with excluded minors and excluded topological minors.2. Notation, terminology, and background material
Let P be a poset. A family R = { L , . . . , L d } of linear extensions of P is called a realizer of P when the following holds: x (cid:54) y in P if and only if x (cid:54) y in all L , . . . , L d . Thus dim( P ) is theleast positive integer d such that P has a realizer of size d . Accordingly, to establish an upperbound of the form dim( P ) (cid:54) d , the most natural approach is simply to construct a realizer ofsize d for P . However, in recent papers [8, 9, 12, 13, 14, 15, 16, 21, 23, 28, 30], another approachhas been taken. Let Inc( P ) denote the set of ordered incomparable pairs of P . Clearly, a family R of linear extensions of P is a realizer of P if and only if for every ( x, y ) ∈ Inc( P ), there is L ∈ R with x > y in L . In this case, we say that L reverses the incomparable pair ( x, y ). Moregenerally, when S is a set of incomparable pairs of P , a linear extension L reverses S when x > y in L for every ( x, y ) ∈ S . A set S ⊆ Inc( P ) is reversible when there is a linear extension L of P that reverses S , and a family R of linear extensions reverses S when for every ( x, y ) ∈ S , thereis L ∈ R that reverses ( x, y ). With these ideas in hand, when S ⊆ Inc( P ), we can define the dimension of S , denoted by dim( S ), as the least positive integer d for which there is a family of d linear extensions of P that reverses S . Clearly, dim( P ) = dim(Inc( P )), so we can also say thatdim( P ) is the least positive integer d for which there is a partition of Inc( P ) into d reversible sets.An indexed family { ( x α , y α ) } sα =1 of incomparable pairs of P with s (cid:62) alternatingcycle when x α (cid:54) y α +1 in P for every index α considered cyclically in { , . . . , s } (that is, y s +1 = y ).An alternating cycle is strict when there are no other comparabilities, that is, x i (cid:54) y j in P if and only if j ≡ i + 1 (mod s ). The following elementary lemma, proved in [27], provides aconvenient test to determine whether a subset of Inc( P ) is reversible. Lemma 2.1. If P is a poset and S ⊆ Inc( P ) , then the following statements are equivalent: (1) S is not reversible; D. M. HOWARD, N. STREIB, W. T. TROTTER, B. WALCZAK, AND R. WANG (2) S contains an alternating cycle; (3) S contains a strict alternating cycle. A typical approach to show that the set Inc( P ) can be partitioned into d reversible sets is bydefining a d -coloring of the pairs in Inc( P ) with the property that no (strict) alternating cycle ismonochromatic. However, the rules for assigning colors can be quite complicated, and that willcertainly be the case here.3. Posets that exclude two long incomparable chains
In this section, we present some general considerations on posets excluding two long incompa-rable chains. If a poset P excludes + , then P is a chain, so dim( P ) = 1. For the rest of thissection, we fix an integer k (cid:62) P that excludes k + k . We make no assumption onthe structure of the cover graph of P .Let h denote the height of P , and let C = { c < · · · < c h } be a chain in P of size h . For eachpoint z ∈ P − C , define integers dn( z ) and up( z ) as follows:dn( z ) = z is incomparable with c in P , i otherwise, where i is greatest in { , . . . , h } such that z > c i in P ;up( z ) = h + 1 if z is incomparable with c h in P , j otherwise, where j is least in { , . . . , h } such that z < c j in P .Note that 0 (cid:54) dn( z ) (cid:54) h − (cid:54) up( z ) (cid:54) h + 1 for every point z ∈ P − C , by maximalityof the chain C . DefineDn( i ) = { z ∈ P − C : dn( z ) = i } for 0 (cid:54) i (cid:54) h − j ) = { z ∈ P − C : up( z ) = j } for 2 (cid:54) j (cid:54) h + 1. Lemma 3.1.
For (cid:54) i (cid:54) h − , the subposet Dn( i ) of P is convex and has height at most k − . More generally, for (cid:54) i (cid:54) i + m (cid:54) h − , the subposet Dn( i, m ) of P defined by Dn( i, m ) = i + m [ α = i Dn( α ) ∪ { c i +1 , . . . , c i + m } is convex and has height at most m + 2 k − .Dually, for (cid:54) j (cid:54) h + 1 , the subposet Up( j ) of P is convex and has height at most k − .More generally, for (cid:54) j (cid:54) j + m (cid:54) h + 1 , the subposet Up( j, m ) of P defined by Up( j, m ) = j + m [ α = j Up( α ) ∪ { c j , . . . , c j + m − } is convex and has height at most m + 2 k − .Proof. We only show the proof of the first part, as the second is dual. It is clear that thesubposet Dn( i, m ) is convex. The fact that C is a maximum chain implies that the height ofDn( i, m ) is at most h − i , so the desired inequality follows if h − i (cid:54) m + 2 k −
2. Suppose h (cid:62) i + m + 2 k − (cid:62) i + m + k . Let Q = { z ∈ Dn( i, m ) : up( z ) (cid:54) c i + m + k } . The fact that C is amaximum chain forces the height of Q to be at most m + k −
1. Furthermore, the height of thesubposet Dn( i, m ) − Q is at most k −
1, because all points of Dn( i, m ) − Q are incomparablewith the k -element chain { c i + m +1 < · · · < c i + m + k } . Hence the height of Dn( i, m ) is at most( m + k −
1) + ( k −
1) = m + 2 k − (cid:3) IMENSION OF POSETS WITH PLANAR COVER GRAPHS EXCL. TWO LONG INCOMP. CHAINS 5
For 0 (cid:54) i (cid:54) h , define Ar( i, i + 1) = { z ∈ P − C : dn( z ) (cid:54) i and up( z ) (cid:62) i + 1 } . Here, Arstands for “around”. Lemma 3.2.
For (cid:54) i (cid:54) h , the subposet Ar( i, i +1) of P is convex and has height at most k − .Proof. It is clear that the subposet Ar( i, i +1) is convex. Let Q = { z ∈ Ar( i, i +1) : up( z ) (cid:54) i + k } .Thus Q ⊆ Up( i + 1) ∪ · · · ∪ Up( i + k ) ⊆ Up( i + 1 , k − Q is at most ( k −
1) + 2 k − k −
3. Furthermore, the height of the subposet Ar( i, i + 1) − Q is at most k −
1, because all points of Ar( i, i + 1) − Q are incomparable with the k -element chain { c i +1 < · · · < c i + k } . Hence the height of Ar( i, i + 1) is at most (3 k −
3) + ( k −
1) = 4 k − (cid:3) Lemma 3.3.
Let z, w ∈ P − C , dn( z ) < up( w ) , and w (cid:54) z in P . If C is a chain in P − C with w the least element and z the greatest element, then | C | (cid:54) k − .Proof. We have z, w ∈ Ar(dn( z ) , dn( z ) + 1), which implies C ⊆ Ar(dn( z ) , dn( z ) + 1). We applyLemma 3.2 to conclude that | C | (cid:54) k − (cid:3) Every incomparable pair ( x, y ) of P satisfies dn( y ) < up( x ). We call an incomparable pair( x, y ) of P dangerous if dn( x ) < dn( y ) < up( x ) < up( y ) and safe otherwise. Lemma 3.4. If d is a positive integer such that every convex subposet Q of P of height atmost k − satisfies dim( Q ) (cid:54) d , then there is a set of at most d linear extensions of P thatreverses all the safe incomparable pairs of P .Proof. It follows from Lemma 3.1 that dim(Dn( i )) (cid:54) d for 0 (cid:54) i (cid:54) h − j )) (cid:54) d for 2 (cid:54) j (cid:54) h + 1. First, consider d linear extensions of P that • have block form Dn(0) < c < Dn(1) < · · · < c h − < Dn( h − < c h , • induce d linear extensions of Dn( i ) witnessing dim(Dn( i )) (cid:54) d , for 0 (cid:54) i (cid:54) h − x, y ) of P suchthat dn( x ) > dn( y ), and by the second condition—all incomparable pairs ( x, y ) of P such thatdn( x ) = dn( y ). Then, consider d more linear extensions of P that • have block form c < Up(2) < c < · · · < Up( h ) < c h < Up( h + 1), • induce d linear extensions of Up( i ) witnessing dim(Up( i )) (cid:54) d , for 2 (cid:54) i (cid:54) h + 1.By the first condition, these linear extensions reverse all incomparable pairs ( x, y ) of P suchthat up( x ) > up( y ), and by the second condition—all incomparable pairs ( x, y ) of P such thatup( x ) = up( y ). We conclude that an incomparable pair ( x, y ) of P is reversed by some of the 2 d linear extensions unless dn( x ) < dn( y ) < up( x ) < up( y ), that is, the pair ( x, y ) is dangerous. (cid:3) In view of Lemma 3.4, we can focus on reversing only the dangerous incomparable pairs whenattempting for a bound on dim( P ). This is the starting point of the proof of Theorem 1.4 in thenext sections. We conclude this section with two results that will not be used further in thepaper: one asserting that dim( P ) = O ( h ) whenever the convex subposets of P with boundedheight have bounded dimension, and the other asserting that linear dependence on h is necessary. Proposition 3.5. If d is a positive integer such that every convex subposet Q of P of heightat most k − satisfies dim( Q ) (cid:54) d , then dim( P ) (cid:54) ( h + 1) d .Proof. We apply Lemma 3.4 to reverse all the safe incomparable pairs of P using at most 2 d linearextensions. For every dangerous incomparable pair ( x, y ) of P , we have x, y ∈ Ar(dn( y ) , dn( y )+1),where 1 (cid:54) dn( y ) (cid:54) h −
1. For 1 (cid:54) i (cid:54) h −
1, it follows from Lemma 3.2 that dim(Ar( i, i +1)) (cid:54) d ,and we can extend any linear extension of Ar( i, i + 1) witnessing the dimension to a linearextension of P . This way, we obtain a set of at most ( h − d linear extensions of P reversingall the dangerous incomparable pairs of P , and the proof is complete. (cid:3) D. M. HOWARD, N. STREIB, W. T. TROTTER, B. WALCZAK, AND R. WANG
Proposition 3.6.
For every positive integer n , there is a poset P excluding + such that dim( P ) (cid:62) n , h ( P ) = n + 1 , and every convex subposet Q of P satisfies dim( Q ) (cid:54) h ( Q ) + 1 ,where h ( Q ) denotes the height of Q .Proof. The poset P consists of points a , . . . , a n , b , . . . , b n , c , . . . , c n with the following coverrelations: c < · · · < c n , a i < c i and c i − < b i for 1 (cid:54) i (cid:54) n , and a i < b j for 1 (cid:54) j < i (cid:54) n . Everychain in P of size at least 3 contains a point of the form c i , so P excludes + . The subposetof P induced on a , . . . , a n , b , . . . , b n is isomorphic to the standard example S n , so dim( P ) (cid:62) n .Every convex subposet of P of height 1 is an antichain and therefore has dimension at most 2.Now, let h (cid:62)
2, and let Q be a convex subposet of P of height h . To prove dim( Q ) (cid:54) h +1, we canassume without loss of generality that Q is a maximal subposet of P of height h . It follows that Q is comprised of points a i +1 , . . . , a n , b , . . . , b i + h − , c i , . . . , c i + h − for some i ∈ { , . . . , n − h + 1 } .It is easy to check that the following h + 1 linear extensions of Q form a realizer of Q : a i +1 < · · · < a n < b i < · · · < b < c i < b i +1 < c i +1 < · · · < b i + h − < c i + h − ,c i < a i +1 < c i +1 < · · · < a i + j − < c i + j − < a i + j +1 < · · · < a n < b i + j < a i + j < b < · · · < b i + j − < c i + j < b i + j +1 < c i + j +1 < · · · < b i + h − < c i + h − for 1 (cid:54) j < h,c i < a i +1 < c i +1 < · · · < a i + h − < c i + h − < a n < · · · < a i + h < b < · · · < b i +1 . (cid:3) Proof of the main theorem
For the proof of Theorem 1.4, we fix a poset P that excludes k + k , where k (cid:62)
2, and has aplanar cover graph, and we attempt to partition the set Inc( P ) of incomparable pairs of P into abounded number of reversible subsets, where the bound depends only on k . Following the notationand terminology of the preceding section, let h be the height of P , and let C = { c < · · · < c h } be a chain in P of size h . We use operators Dn, Up, and Ar as in the preceding section to denoteappropriate convex subposets of P − C . We use operators dn and up in a different way than inthe preceding section, namely, to refer to points of the chain C rather than integer numbers: • for z ∈ P , we let dn( z ) = c i if c i (cid:54) z and i is greatest in { , . . . , h } with this property; • for z ∈ P , we let up( z ) = c i if z (cid:54) c i and i is least in { , . . . , h } with this property.We can write dn( z ) or up( z ) only for points z ∈ P for which the respective point in C exists.By Theorem 1.3, there is an integer d such that every poset P with a planar cover graph andwith height at most 4 k − P ) (cid:54) d . It follows that every convex subposet of P ofheight at most 4 k − d . For 1 (cid:54) i (cid:54) h −
1, in particular, Lemma 3.2yields dim(Ar( i, i + 1)) (cid:54) d , so there is a coloring φ i : Inc(Ar( i, i + 1)) → { , . . . , d } such thatfor each color γ ∈ { , . . . , d } , the set of incomparable pairs of Ar( i, i + 1) that are assigned color γ by φ i is reversible. We fix the integer d and the colorings φ i for 1 (cid:54) i (cid:54) h − P using at most 2 d linearextensions, and the remaining challenge is to reverse the dangerous incomparable pairs of P . Forevery dangerous incomparable pair ( a, b ) of P , we have c < b and a < c h in P , so the pointsdn( b ) and up( a ) of the chain C are defined. For any set S of dangerous incomparable pairs of P ,let A ( S ) denote the set of points a ∈ P for which there is a point b ∈ P with ( a, b ) ∈ S , and let B ( S ) denote the set of points b ∈ P for which there is a point a ∈ P with ( a, b ) ∈ S .Recall from the preceding section that for every dangerous incomparable pair ( a, b ) of P , wehave a, b ∈ Ar( i, i + 1) ∩ · · · ∩
Ar( j − , j ), where c i = dn( b ) and c j = up( a ). In particular, thedangerous incomparable pairs ( a, b ) of P with dn( b ) = c belong to Inc(Ar(1 , IMENSION OF POSETS WITH PLANAR COVER GRAPHS EXCL. TWO LONG INCOMP. CHAINS 7 be reversed using d linear extensions. Similarly, the dangerous incomparable pairs ( a, b ) of P with up( a ) = c h can be reversed using d linear extensions.Let S be the set of dangerous incomparable pairs ( a, b ) of P with dn( b ) > c and up( a ) < c h in P . It remains to partition the set S into a bounded number of reversible subsets. By convention,we will write S only to denote some subset of the set S , we will write a , a , a α , etc. only todenote a point from A ( S ), and we will write b , b , b α , etc. only to denote a point from B ( S ).For a set S ⊆ S , we call a pair ( a, b ) ∈ S left-safe with respect to S if there is no point b ∈ B ( S ) with a (cid:54) b and dn( b ) < dn( b ) in P . The following lemma plays an important role inour argument. Lemma 4.1.
For a set S ⊆ S , if S is the set of pairs in S that are left-safe with respect to S ,then dim( S ) (cid:54) d .Proof. For a color γ ∈ { , . . . , d } , let S ( γ ) be the subset of S consisting of all pairs ( a, b ) ∈ S such that φ i ( a, b ) = γ , where c i = dn( b ). It is enough to prove that the set S ( γ ) is reversiblefor every γ ∈ { , . . . , d } . Suppose to the contrary that the set S ( γ ) is not reversible forsome γ ∈ { , . . . , d } . This means that S ( γ ) contains an alternating cycle { ( a α , b α ) } sα =1 , where s (cid:62)
2. For every α ∈ { , . . . , s } , we have a α (cid:54) b α +1 and therefore dn( b α ) (cid:54) dn( b α +1 ) in P ,because the pair ( a α , b α ) is left-safe with respect to S . Since the latter inequality holds forevery α ∈ { , . . . , s } , there is a point c i ∈ C with 1 (cid:54) i (cid:54) h − c i = dn( b α ) forevery α ∈ { , . . . , s } . This implies that { ( a α , b α ) } sα =1 is a monochromatic alternating cycle inInc(Ar( i, i + 1)), which is a contradiction. (cid:3) Let G denote the cover graph of P , which is a planar graph. We fix a plane straight-linedrawing of G , that is, a drawing of G in the plane using non-crossing straight-line segments foredges. We assume, without loss of generality, that the least point c of the maximum chain C lies on the outer face of the drawing.A witnessing path for a pair ( x, y ) with x (cid:54) y in P is a path u · · · u r in G such that x = u < · · · < u r = y in P (in particular, u i is covered by u i +1 in P for 0 (cid:54) i (cid:54) r − x, y ) with x (cid:54) y in P has at least one witnessing path. For thepurpose of our proof, it is convenient to fix one witnessing path, to be denoted by W ( x, y ), foreach pair ( x, y ) with x (cid:54) y in P . However, we need to choose the paths W ( x, y ) in a consistentway, which is achieved in the following lemma. Lemma 4.2.
There is a function ( x, y ) W ( x, y ) that maps each pair ( x, y ) with x (cid:54) y in P to a witnessing path W ( x, y ) for ( x, y ) in such a way that the following holds: (1) if x (cid:54) c (cid:54) y in P and c ∈ C , then W ( x, y ) passes through c ; in particular, if up( x ) and dn( y ) are defined and up( x ) (cid:54) dn( y ) in P , then W ( x, y ) passes through up( x ) and dn( y ) ; (2) if x (cid:54) x (cid:54) y (cid:54) y in P and W ( x , y ) passes through x and y , then W ( x , y ) is thesubpath of W ( x , y ) from x to y .Proof. Assume some (arbitrary) total order ≺ on the points of P . The order ≺ extends naturallyto a lexicographic total order ≺ lex on finite sequences of points of P as follows: • the empty sequence is the least element in ≺ lex ; • for any two non-empty sequences u · · · u r and v · · · v s of points of P , we have u · · · u r ≺ lex v · · · v s if and only if u < v or u = v and u · · · u r ≺ lex v · · · v s .For every pair ( x, y ) with x (cid:54) y in P , let W ( x, y ) be the ≺ lex -minimum witnessing path amongall witnessing paths for ( x, y ) passing through all points c ∈ C with x (cid:54) c (cid:54) y in P . It is clearthat the paths W ( x, y ) so defined satisfy both conditions of the lemma. (cid:3) D. M. HOWARD, N. STREIB, W. T. TROTTER, B. WALCZAK, AND R. WANG
The length of a witnessing path W ( x, y ), denoted by k W ( x, y ) k , is the number of edges in W ( x, y ). Lemmas 3.1 and 3.2 imply the following bounds on the lengths of witnessing paths: • k W (dn( z ) , z ) k (cid:54) k − z ∈ P , • k W ( z, up( z )) k (cid:54) k − z ∈ P , • k W ( z, w ) k (cid:54) k − z, w ∈ P − C with dn( w ) < up( z ).For i ∈ { , . . . , h − } , we classify every edge of the form c i z of G , where z / ∈ C , as a • left-edge if the edges c i c i − , c i z , c i c i +1 occur in this order clockwise around c i , • right-edge if the edges c i c i − , c i z , c i c i +1 occur in this order counterclockwise around c i .Since c (cid:54) dn( b ) < up( a ) (cid:54) c h − in P for any ( a, b ) ∈ S , the above yields a partition of S intofour classes S LL , S LR , S RL , and S RR according to how the last edge of W ( a, up( a )) and the firstedge of W (dn( b ) , b ) are classified. That is, for every ( a, b ) ∈ S , we have • ( a, b ) ∈ S LL ∪ S LR if the last edge of W ( a, up( a )) is a left-edge, • ( a, b ) ∈ S RL ∪ S RR if the last edge of W ( a, up( a )) is a right-edge, • ( a, b ) ∈ S LL ∪ S RL if the first edge of W (dn( b ) , b ) is a left-edge, • ( a, b ) ∈ S LR ∪ S RR if the first edge of W (dn( b ) , b ) is a right-edge.To complete the proof of Theorem 1.4, we will establish the following: • dim( S LL ) = O ( k d ) and dim( S RR ) = O ( k d ), in the next section; • dim( S LR ) = O ( k + k d ) and dim( S RL ) = O ( k + k d ), in Section 6.This allows us to conclude that dim( S ) = O ( k d ) and thus dim( P ) = O ( k d ).5. Same-side dangerous pairs
In this section, we show that dim( S LL ) = O ( k d ) and dim( S RR ) = O ( k d ). We present theargument only for S LL , and the argument for S RR is symmetric. To simplify the notation usedin this portion of the proof, we (temporarily) set S = S LL .Recall that c < b in P for every b ∈ B ( S ), as S contains only dangerous pairs with dn( b ) > c in P . Furthermore, the choice of the witnessing paths guarantees that • for any b ∈ B ( S ), the common part of W ( c , b ) with W ( c , c h ) is W ( c , dn( b )), • for any b, b ∈ B ( S ), the common part of W ( c , b ) and W ( c , b ) is W ( c , z ) for some z ∈ P .Therefore, the union of W ( c , c h ) and the witnessing paths W ( c , b ) over all b ∈ B ( S ) forms atree, which we call the basic tree and denote by BT. The points, edges, and paths in BT arecalled basic points , basic edges , and basic paths . For any two basic points u and v , let BT( u, v )denote the unique path in BT (made of basic points and basic edges) between u and v .Let ≺ be the total order on the basic points determined by carrying out a depth-first searchof BT from c that processes the children of each node in the clockwise order and sets u ≺ v when u is visited before v for the first time. In other words, for any two basic points u and v , if W ( c , z ) is the common part of W ( c , u ) and W ( c , v ), c = z = v , and either u ∈ W ( c , v ) orthe basic paths BT( z, c ) (which is the reverse of W ( c , z )), W ( z, u ), and W ( z, v ) go out of z inthis order clockwise, then we declare u ≺ v . It follows that u ≺ v for any two basic points u and v such that dn( u ) < dn( v ) in P . The greatest point in this order is c h . Figure 2 illustrates howthe basic tree might appear.Let a ∈ A ( S ). Following [23], we call a basic point v special for a if the following holds: • a (cid:54) v in P , • a (cid:54) u in P whenever u ∈ W ( c , v ) and u = v . IMENSION OF POSETS WITH PLANAR COVER GRAPHS EXCL. TWO LONG INCOMP. CHAINS 9 c c c c c c c c c c c c u u u u a bx yt R ( t, x, y ) Figure 2.
Illustration for the concepts introduced in Section 5: arrows on thecover graph edges point according to the increasing direction of P ; the basic treeis marked by empty points and double edges; Spec( a ) = { u ≺ u ≺ u ≺ t ≺ x ≺ u ≺ y ≺ c } ; Spec ( a ) = { u ≺ x ≺ y ≺ c } .Let Spec( a ) denote the set of all basic points v that are special for a (see Figure 2). Any twodistinct points v, v ∈ Spec( a ) satisfy v / ∈ W ( c , v ) and v / ∈ W ( c , v ), although they may becomparable in P . The set Spec( a ) inherits the order ≺ from BT. Since up( a ) ∈ Spec( a ), wehave Spec( a ) = ∅ . In fact, up( a ) is the greatest point in the order ≺ on Spec( a ).The following straightforward property is stated for emphasis. Observation 5.1. If a ∈ A ( S ) , b ∈ B ( S ) , and a (cid:54) b in P , then there is v ∈ Spec( a ) with v ∈ W ( c , b ) and v (cid:54) b in P . Now, suppose a ∈ A ( S ), x, y ∈ Spec( a ), and x ≺ y in BT. Every point t ∈ P such that a (cid:54) t (cid:54) x, y in P and t is the only common point of W ( t, x ) and W ( t, y ) gives rise to a regionin the plane, denoted by R ( t, x, y ), whose boundary is the simple closed curve formed by thepaths BT( x, y ), W ( t, x ), and W ( t, y ). The boundary of R ( t, x, y ) traversed counterclockwisegoes from t to x along W ( t, x ), then to y along BT( x, y ), and then back to t along W ( t, y ) inthe reverse direction (see Figure 2). It is possible that t = x or t = y . If a = t , then a lies eitherin the interior or in the exterior of R ( t, x, y ), but not on the boundary. Any region of the form R ( t, x, y ) where x, y ∈ Spec( a ) and a (cid:54) t in P is called an a -region . (For such an a -region, it is not required that t lies on the paths W ( a, x ) and W ( a, y ).)Let T denote the subset of S consisting of all pairs that are left-safe with respect to S . ByLemma 4.1, we have dim( T ) (cid:54) d . To complete the proof, we will show that dim( S − T ) (cid:54) (cid:0) k − (cid:1) (4 k − d by partitioning S − T into reversible subsets of the form T ( h , h , n, γ ), wherethe four parameters are integers with 0 (cid:54) h < h (cid:54) k −
2, 0 (cid:54) n (cid:54) k −
5, and 1 (cid:54) γ (cid:54) d .Membership in these sets will be determined in two stages. In the first stage, we will partition S − T into subsets of the form T ( h , h ) where 0 (cid:54) h < h (cid:54) k −
2. In the second stage, foreach pair ( h , h ) with 0 (cid:54) h < h (cid:54) k −
2, we will further partition T ( h , h ) into reversiblesubsets of the form T ( h , h , n, γ ), where 0 (cid:54) n (cid:54) k − (cid:54) γ (cid:54) d .To describe the first partition, we need some more definitions. For a basic point v , let h ( v ) = k W (dn( v ) , v ) k . We have 0 (cid:54) h ( v ) (cid:54) k − v , and we have h ( v ) = 0if and only if v ∈ C . For a point a ∈ A ( S − T ), letSpec ( a ) = { u ∈ Spec( a ) : for every v ∈ Spec( a ), if v ≺ u in BT, then h ( v ) > h ( u ) } (see Figure 2). The set Spec ( a ) inherits the order ≺ from Spec( a ). The sequence of numbers h ( v ) for the points v ∈ Spec ( a ) considered in the order ≺ is strictly decreasing. The first pointin Spec ( a ) is the first point in Spec( a ), and the last point in Spec ( a ) is up( a ). Now, let ( a, b ) ∈ S − T . Since ( a, b ) is not left-safe with respect to S , there is a point u ∈ Spec( a ) with dn( u ) < dn( b ) in P . Consequently, there is a point u ∈ Spec ( a ) withdn( u ) < dn( b ) in P and thus u ≺ b in BT. We also have b ≺ up( a ) ∈ Spec ( a ) and b / ∈ Spec ( a ).If follows that there are two points x, y ∈ Spec ( a ) consecutive in the order ≺ on Spec ( a ) suchthat x ≺ b ≺ y in BT. Let h = h ( y ) and h = h ( x ), so that 0 (cid:54) h < h (cid:54) k −
2. We put thepair ( a, b ) to the set T ( h , h ) of the first partition.To complete the proof, it remains to show that dim( T ( h , h )) (cid:54) (4 k − d for each pair ofintegers ( h , h ) with 0 (cid:54) h < h (cid:54) k −
2. To this end, we fix an arbitrary pair ( h , h ) of thisform and show that dim( T ( h , h )) (cid:54) (4 k − d by explaining how T ( h , h ) can be partitionedinto reversible sets of the form T ( h , h , n, γ ), where 0 (cid:54) n (cid:54) k − (cid:54) γ (cid:54) d . This taskwill require some preliminary work.In the reasoning used above to put a pair ( a, b ) in T ( h , h ), the points x and y have beendefined depending on both a and b . Now, however, for any point a ∈ A ( T ( h , h )), there isa unique point y ∈ Spec ( a ) with h ( y ) = h , and there is a unique point x ∈ Spec ( a ) with h ( x ) = h . Let y ( a ) and x ( a ) denote these points, respectively, for any a ∈ A ( T ( h , h )). Lemma 5.2.
Let ( a, b ) , ( a , b ) ∈ T ( h , h ) be such that a (cid:54) b and dn( y ( a )) < dn( y ( a )) in P . Then, for every a -region of the form R ( t, x ( a ) , y ( a )) , there is an a -region of the form R ( t , x ( a ) , y ( a )) such that k W ( t , x ( a )) k < k W ( t, x ( a )) k .Proof. Fix an a -region R ( t, x ( a ) , y ( a )), where a (cid:54) t in P . For simplicity of notation, let x = x ( a ), y = y ( a ), R = R ( t, x ( a ) , y ( a )), x = x ( a ), and y = y ( a ). We prove the lemma in several steps,establishing the following claims:(1) y lies in the exterior of R ;(2) b lies in the interior or on the boundary of R ;(3) a (cid:54) y in P ;(4) if a (cid:54) u in P and u ∈ W ( c , x ), then u = x ;(5) a lies in the interior of R ;(6) there is a point t on W ( t, x ) such that a (cid:54) t (cid:54) x, y in P , t = t , and t is the only commonpoint of W ( t , x ) and W ( t , y );(7) x = x .The conclusion of the lemma then follows directly from (6) and (7).For the proof of (1), suppose y does not lie in the exterior of R . Since dn( y ) < dn( y ) in P , thefirst edge of the path W (dn( y ) , y ) is part of the chain C and lies in the exterior of R . Therefore,the path W (dn( y ) , y ) crosses the boundary of R at some point u other than dn( y ). It followsthat u (cid:54) x or u (cid:54) y in P , which contradicts the fact that dn( x ) (cid:54) dn( y ) < dn( y ) = dn( u ) in P .For the proof of (2), suppose b lies in the exterior of R . Since x ≺ b ≺ y in BT, the basic path W (dn( b ) , b ) enters the interior of R with the first edge that is not common with W (dn( x ) , x ) nor W (dn( y ) , y ). Therefore, it must exit the interior of R through a point u on W ( t, x ) or W ( t, y ).It follows that a (cid:54) u (cid:54) b in P , which is a contradiction.For the proof of (3), suppose a (cid:54) y in P . It follows that a has a special point u on the basicpath W ( c , y ). We have h ( u ) (cid:54) h ( y ) = h = h ( y ). However, we also have dn( u ) (cid:54) dn( y ) < dn( y ) in P and thus u ≺ y in BT. This is a contradiction with y ∈ Spec ( a ).For the proof of (4), suppose a (cid:54) u in P , u ∈ W ( c , x ), and u = x . Assume without loss ofgenerality that u is a special point for a . We have h ( u ) < h ( x ) = h = h ( x ) and thus x ≺ u inBT. We also have dn( u ) (cid:54) dn( y ) < dn( y ) in P and thus u ≺ y in BT, which implies h ( u ) > h ( y ),as y ∈ Spec ( a ). However, since x and y are consecutive in the order ≺ on Spec ( a ), no point u ∈ Spec( a ) with x ≺ u ≺ y in BT can satisfy h ( x ) > h ( u ) > h ( y ), which is a contradiction. IMENSION OF POSETS WITH PLANAR COVER GRAPHS EXCL. TWO LONG INCOMP. CHAINS 11
For the proof of (5), suppose a does not lie in the interior of R . By (2), b is not in the exteriorof R , so the path W ( a , b ) intersects the boundary of R at some point u . Since a (cid:54) u in P , itfollows from (3) and (4) that u lies on W ( t, x ). Thus a (cid:54) t (cid:54) u (cid:54) b in P , which is a contradiction.By (5) and (1), a is in the interior and y is in the exterior of R , so the path W ( a , y ) crossesthe boundary of R . For the proof of (6), let t be the last common point of W ( a , y ) with theboundary of R in the order along W ( a , y ). Since a (cid:54) t in P , it follows from (3) and (4) that t lies on W ( t, x ) and t = t . Furthermore, it follows from the choice of t that it is the onlycommon point of W ( t , x ) and W ( t , y ). Therefore, t satisfies all the conditions of (6).It remains to prove (7). Suppose x = x . The fact that a (cid:54) t (cid:54) x in P and (4) imply that x ∈ Spec( a ). Since h ( x ) = h = h ( x ) and x ∈ Spec ( a ), we have x ≺ x in BT. Therefore,the first edge of the path W ( c , x ) that is not common with W ( c , x ) lies in the exterior of R .If x is not in the exterior of R , then the path W ( c , x ) crosses W ( t, x ) or W ( t, y ) and thus a (cid:54) t (cid:54) x in P . If x is not in the interior of R , then the path W ( a , x ) crosses the boundaryof R at some point u ; again, since a (cid:54) u in P , it follows from (3) and (4) that u lies on W ( t, x )and thus a (cid:54) t (cid:54) u (cid:54) x in P . In either case, we have concluded that a (cid:54) x in P . Consequently,there is a point v ∈ Spec( a ) on the path W ( c , x ). It follows that v ≺ x ≺ x in BT and h ( v ) (cid:54) h ( x ) = h ( x ), which is a contradiction to x ∈ Spec ( a ). Thus x = x . (cid:3) For incomparable pairs ( a, b ) , ( a , b ) ∈ T ( h , h ), let ( a, b ) → ( a , b ) denote that a (cid:54) b anddn( y ( a )) < dn( y ( a )) in P . For a pair ( a, b ) ∈ T ( h , h ), let ‘ ( a, b ) denote the greatest integer s forwhich there is a sequence { ( a α , b α ) } sα =0 ⊆ T ( h , h ) such that ( a , b ) → · · · → ( a s , b s ) = ( a, b ). Lemma 5.3.
For every incomparable pair ( a, b ) ∈ T ( h , h ) , the following holds: (1) 0 (cid:54) ‘ ( a, b ) (cid:54) k − ; (2) if ( a, b ) → ( a , b ) ∈ T ( h , h ) , then ‘ ( a, b ) < ‘ ( a , b ) .Proof. Fix a sequence { ( a α , b α ) } sα =0 ⊆ T ( h , h ) such that ( a , b ) → · · · → ( a s , b s ) = ( a, b ),where s = ‘ ( a, b ). To see (1), choose any a -region of the form R ( t , x ( a ) , y ( a )), and applyLemma 5.2 repeatedly to obtain a sequence of regions {R ( t α , x ( a α ) , y ( a α )) } sα =0 such that k W ( t , x ( a )) k > · · · > k W ( t s , x ( a s )) k (cid:62)
0; these inequalities are possible only when s (cid:54) k W ( t , x ( a )) k (cid:54) k −
5, where the latter inequality follows from the fact that dn( x ( a )) < up( a ) (cid:54) up( t ). To see (2), set ( a s +1 , b s +1 ) = ( a , b ) and observe that the sequence { ( a α , b α ) } s +1 α =0 witnesses ‘ ( a , b ) (cid:62) s + 1. (cid:3) We partition the set T ( h , h ) into subsets of the form T ( h , h , n, γ ), putting every pair( a, b ) ∈ T ( h , h ) into the set T ( h , h , n, γ ) such that n = ‘ ( a, b ) and γ is determined as follows: • if h >
0, then dn( b ) (cid:54) dn( y ( a )) < up( a ) in P , so ( a, b ) is an incomparable pair of Ar( i, i + 1),where c i = dn( y ( a )), and we let γ = φ i ( a, b ); • if h = 0, then y ( a ) = up( a ), so ( a, b ) is an incomparable pair of Ar( j − , j ), where c j = up( a ),and we let γ = φ j − ( a, b ).It follows that 0 (cid:54) n (cid:54) k − (cid:54) γ (cid:54) d . To complete the proof, itremains to show the following. Lemma 5.4.
Every set T ( h , h , n, γ ) is reversible.Proof. Suppose not. Pick an alternating cycle { ( a α , b α ) } sα =1 contained in T ( h , h , n, γ ), where s (cid:62)
2. For 1 (cid:54) α (cid:54) s , since ‘ ( a α , b α ) = n = ‘ ( a α +1 , b α +1 ), it follows from Lemma 5.3 (2) that( a α +1 , b α +1 ) ( a α , b α ). This and a α (cid:54) b α +1 in P yield dn( y ( a α )) (cid:54) dn( y ( a α +1 )) in P , for1 (cid:54) α (cid:54) s . This implies that there is c ∈ C with dn( y ( a α )) = c for 1 (cid:54) α (cid:54) s . If h > c i = c , then { ( a α , b α ) } sα =1 is an alternating cycle in Ar( i, i + 1), and φ i ( a α , b α ) = γ for 1 (cid:54) α (cid:54) s , which is a contradiction. If h = 0 and c j = c , then { ( a α , b α ) } sα =1 is an alternating cycle inAr( j − , j ), and φ j − ( a α , b α ) = γ for 1 (cid:54) α (cid:54) s , which is again a contradiction. (cid:3) We have partitioned S into O ( k d ) reversible subsets of the form T ( h , h , n, γ ) with 0 (cid:54) h < h (cid:54) k −
2, 0 (cid:54) n (cid:54) k −
5, and 1 (cid:54) γ (cid:54) d , thus proving that dim( S ) = O ( k d ).6. Opposite-side dangerous pairs
In this section, we show that dim( S LR ) = O ( k d + k ) and dim( S RL ) = O ( k d + k ). Wepresent the argument only for S RL , and the argument for S LR is symmetric. Recall that the set S RL contains only dangerous incomparable pairs ( a, b ) of P with dn( b ) > c and up( a ) < c h in P . We begin by setting S = S RL . As the argument proceeds, the meaning of S changes, butthe “new” set S is always a subset of the “old” set S . Each time the meaning of S changes, thetarget upper bound on dim( S ) is adjusted accordingly.In the argument given thus far, our main emphasis has been on classifying incomparable pairs.Now, we want to pay attention to comparable pairs. When a ∈ A ( S ), b ∈ B ( S ), and a (cid:54) b in P ,we call ( a, b ) a strong comparable pair if up( a ) (cid:54) dn( b ) in P , and we call ( a, b ) a weak comparablepair if up( a ) > dn( b ) in P . When a comparable pair ( a, b ) is weak, the witnessing path W ( a, b )has length at most 4 k − C . Lemma 6.1.
For every strict alternating cycle { ( a α , b α ) } sα =1 in S , where s (cid:62) , there is atmost one index α ∈ { , . . . , s } such that ( a α , b α +1 ) is a strong comparable pair.Proof. Suppose there are two distinct indices α, β ∈ { , . . . , s } such that the comparable pairs( a α , b α +1 ) and ( a β , b β +1 ) are strong. Without loss of generality, we have up( a α ) (cid:54) up( a β ) in P .It follows that a α < up( a α ) (cid:54) up( a β ) (cid:54) dn( b β +1 ) < b β +1 in P , which contradicts the assumptionthat the alternating cycle { ( a α , b α ) } sα =1 is strict. (cid:3) When T ⊆ S , let Q ( T ) denote the set of weak comparable pairs ( a, b ) such that a ∈ A ( T ) and b ∈ B ( T ). By Lemma 6.1, if Q ( T ) = ∅ , then there is no strict alternating cycle in T and hence T is reversible. In particular, we can assume for the rest of this section that Q ( S ) = ∅ .For every weak comparable pair ( a, b ) ∈ Q ( S ), the choice of the witnessing paths W ( a, up( a ))(which ends with a right-edge), W (dn( b ) , b ) (which starts with a left-edge), and W ( a, b ) guaranteesthe following: • the common part of W ( a, up( a )) and W ( a, b ) is W ( a, v ) for some point v ; • the common part of W (dn( b ) , b ) and W ( a, b ) is W ( w, b ) for some point w .This yields a region D ( a, b ) in the plane whose boundary is a simple closed curve formed by thefollowing four paths (see Figure 3): • W (dn( b ) , up( a )), called the middle of D ( a, b ), • W ( v, up( a )), called the bottom of D ( a, b ), • W (dn( b ) , w ), called the top of D ( a, b ), • W ( v, w ), called the right side of D ( a, b ).The name for W ( v, w ) is justified by the assumption (made at the beginning of the proof) that c lies on the outer face of the drawing of G and thus c h lies inside D ( a, b ).Next, we determine a positive integer n and a sequence of weak comparable pairs { ( a i , b i ) } n − i =1 using the following “greedy” procedure. We start by choosing ( a , b ) ∈ Q ( S ) so as to maximizedn( b ). Then, for i (cid:62)
2, after ( a i − , b i − ) has been determined, we consider all pairs ( a, b ) ∈ Q ( S )such that the region D ( a i − , b i − ) is contained in the interior of the region D ( a, b ), that is,dn( b ) < up( a ) < dn( b i − ) in P and the boundaries of D ( a, b ) and D ( a i − , b i − ) are disjoint. If IMENSION OF POSETS WITH PLANAR COVER GRAPHS EXCL. TWO LONG INCOMP. CHAINS 13 c c c c c c c a b a b D D Figure 3.
Illustration for the concepts used in Section 6: D = D ( a , b ) and D = D ( a , b ).there is no such pair ( a, b ), then we set n = i and the construction terminates. Otherwise, fromall such pairs ( a, b ), we choose ( a i , b i ) so as to maximize dn( b i ) (see Figure 3).For 1 (cid:54) i (cid:54) n −
1, let u i = dn( b i ) and D i = D ( a i , b i ). Furthermore, let u = c h and u n = c .It follows from the construction that u n < u n − < · · · < u < u in P . Define functions M : A ( S ) → { , . . . , n } and N : B ( S ) → { , . . . , n } as follows: • for a ∈ A ( S ), M ( a ) is the least positive integer j such that u j (cid:54) up( a ); • for b ∈ B ( S ), N ( b ) is the least positive integer j such that u j (cid:54) dn( b ).Let q = 8 k − Lemma 6.2. If ( a, b ) ∈ Q ( S ) , then N ( b ) (cid:54) M ( a ) + q .Proof. Let i = M ( a ) and j = N ( b ). The graph G contains a path W from up( a ) to dn( b ) whichis formed by the bottom, the right side, and the top of D ( a, b ). The bottom and the top of D ( a, b ) have length at most 2 k −
2, while the right side has length at most 4 k −
5, so W haslength at most (2 k −
2) + (2 k −
2) + (4 k −
5) = 8 k −
9. In particular, there are at most 8 k − W distinct from up( a ) and dn( b ).The path W starts at the point up( a ), which is in the interior of all D i +1 , . . . , D j − , andends at the point dn( b ), which is in the exterior of all D i +1 , . . . , D j − . Therefore, it mustcross the boundaries of the regions D i +1 , . . . , D j − in j − i − j − i − (cid:54) k −
10, so that j (cid:54) i + 8 k − i + q . (cid:3) For every r ∈ { , . . . , q − } , let S ( r ) = { ( a, b ) ∈ S : M ( a ) ≡ r (mod q ) } . Thus S = S q − r =0 S ( r ).Since q = O ( k ), it suffices to show that dim( S ( r )) = O ( kd + k ) for every r ∈ { , . . . , q − } tocomplete the proof. So we fix a value of r and update the meaning of S by setting S = S ( r ).For every m ∈ { , . . . , n } with m ≡ r (mod q ), let S m = { ( a, b ) ∈ S : M ( a ) = N ( b ) = m } . Lemma 6.3.
We have dim( S ) (cid:54) d + max { dim( S m ) : 1 (cid:54) m (cid:54) n , m ≡ r (mod M ) } .Proof. Let d = max { dim( S m ) : 1 (cid:54) m (cid:54) n, m ≡ r (mod q ) } . For every m ∈ { , . . . , n } with m ≡ r (mod q ), there is a coloring ψ m that assigns a color ψ m ( a, b ) ∈ { d + 1 , . . . , d + d } toevery incomparable pair ( a, b ) ∈ S m so that the set { ( a, b ) ∈ S m : ψ m ( a, b ) = γ } is reversible forevery color γ ∈ { d + 1 , . . . , d + d } .Every pair ( a, b ) ∈ S with N ( b ) > M ( a ) is an incomparable pair of Ar( j − , j ), where c j = u M ( a ) . Recall that φ j − is a coloring of Inc(Ar( j − , j )) that uses colors { , . . . , d } andavoids monochromatic alternating cycles. For every γ ∈ { , . . . , d } , let S ( γ ) be the subset of S consisting of all pairs ( a, b ) such that N ( b ) > M ( a ) and φ j − ( a, b ) = γ , where c j = u M ( a ) . Forevery γ ∈ { d + 1 , . . . , d + d } , let S ( γ ) be the subset of S consisting of all pairs ( a, b ) such that N ( b ) = M ( a ) and ψ M ( a ) ( a, b ) = γ . Thus S = S d + d γ =1 S ( γ ).We claim that the set S ( γ ) is reversible for every γ ∈ { , . . . , d + d } . Suppose not. Let { ( a α , b α ) } sα =1 be an alternating cycle contained in S ( γ ), where s (cid:62)
2. Suppose there is α ∈{ , . . . , s } with M ( a α +1 ) > M ( a α ). Then N ( b α +1 ) (cid:62) M ( a α +1 ) (cid:62) M ( a α ) + q , as M ( a α +1 ) ≡ M ( a α ) (mod q ). It follows that ( a α , b α +1 ) is a weak comparable pair, but it violates Lemma 6.2.This shows that M ( a ) , . . . , M ( a s ) are all equal. Let m = M ( a ) = · · · = M ( a s ). Consequently, { ( a α , b α ) } sα =1 is an alternating cycle either in Ar( j − , j ), when γ ∈ { , . . . , d } and c j = u m , orin S m , when γ ∈ { d + 1 , . . . , d + d } . In both cases, this is a contradiction. (cid:3) It remains to show that dim( S m ) = O ( kd + k ) for 1 (cid:54) m (cid:54) n . So we fix a value of m andupdate the meaning of S by setting S = S m . It follows that • u m < up( a ) < u m − in P for every a ∈ A ( S ), • u m (cid:54) dn( b ) < u m − in P for every b ∈ B ( S ).If m = 1, then let D = ∅ . If m (cid:62)
2, then let D denote the set of points x on the boundary of D m − such that u m − (cid:54) x in P . It follows that D is contained in the union of the bottom and the rightside of D m − with excluded topmost points, and therefore | D | (cid:54) (2 k −
2) + (4 k − − O ( k ). Lemma 6.4.
Every pair ( a, b ) ∈ Q ( S ) satisfies at least one of the following: (1) dn( b ) = u m ; (2) the top of D ( a, b ) intersects D ; (3) the right side of D ( a, b ) intersects D .Proof. Let ( a, b ) ∈ Q ( S ). Suppose dn( b ) = u m , so that u m < dn( b ) < up( a ) < u m − in P . The“greedy” construction of the sequence { ( a i , b i ) } n − i =1 rejected the pair ( a, b ), so m (cid:62) D m − isnot contained in the interior of D ( a, b ). Since the middle of D m − lies in the interior of D ( a, b ),it follows that the top, the bottom, or the right side of D ( a, b ) intersects the boundary of D m − .If the bottom of D ( a, b ) intersects the top of D m − at a point x , then x (cid:54) up( a ) < u m − =dn( b m − ) (cid:54) x in P , which is a contradiction. If the bottom of D ( a, b ) intersects the bottom or theright side of D m − at a point x , then a m − (cid:54) x (cid:54) up( a ) < u m − < up( a m − ) in P , which showsthat up( a ) is a better candidate for up( a m − ). If the top or the right side of D ( a, b ) intersectsthe boundary of D m − at a point x such that u m − (cid:54) x in P , then dn( b ) < u m − (cid:54) x (cid:54) b in P ,which shows that u m − is a better candidate for dn( b ). We conclude that the top or the rightside of D ( a, b ) intersects D . (cid:3) Let R be the subset of the maximum chain C consisting of u m and all points of the form dn( x )such that x ∈ D and u m < dn( x ) < u m − . It follows that | R | (cid:54) | D | = O ( k ). Let S = { ( a, b ) ∈ S : dn( b ) ∈ R } . For every pair ( a, b ) ∈ S with dn( b ) = c i , we have ( a, b ) ∈ Inc(Ar( i, i + 1)) andthus dim( { ( a, b ) ∈ S : dn( b ) = c i } ) (cid:54) d . It follows that dim( S ) (cid:54) | R | · d = O ( kd ). Lemma 6.5.
For every pair ( a, b ) ∈ Q ( S − S ) , the right side of D ( a, b ) intersects D .Proof. Let ( a, b ) ∈ Q ( S − S ). Since b ∈ B ( S − S ), there is a ∈ A ( S − S ) such that ( a , b ) ∈ S − S .If dn( b ) = u m , then ( a , b ) ∈ S (as u m ∈ R ), which is a contradiction. Now, suppose the topof D ( a, b ) intersects D at a point x . Since x ∈ W (dn( b ) , b ), we have dn( b ) = dn( x ) ∈ R , so( a , b ) ∈ S , which is a contradiction. We have excluded the cases (1) and (2) from Lemma 6.4,so the case (3) must hold. (cid:3) We update the meaning of S once again by setting S = S − S , and we prove that dim( S ) = O ( k ). Let X denote the family of subsets X of D that are downward-closed in D , that is, IMENSION OF POSETS WITH PLANAR COVER GRAPHS EXCL. TWO LONG INCOMP. CHAINS 15 such that y ∈ X whenever x ∈ X , y ∈ D , and y (cid:54) x in P . Every nonempty set X ∈ X ischaracterized by the pair of points ( x, y ) such that x is the topmost point of X on the bottomof D m − and y is the topmost point of X on the right side of D m − . It follows that there are atmost (2 k − k −
5) nonempty sets in X , so |X | (cid:54) (2 k − k −
5) + 1 = O ( k ).For a point b ∈ B ( S ), let ↓ b = { x ∈ P : x (cid:54) b in P } . For every b ∈ B ( S ), we have ↓ b ∩ D ∈ X .For every X ∈ X , let S X = { ( a, b ) ∈ S : ↓ b ∩ D = X } . Lemma 6.6. Q ( S X ) = ∅ for every X ∈ X .Proof. Suppose there is a pair ( a, b ) ∈ Q ( S X ). By Lemma 6.5, the right side of D ( a, b ) intersects D , so there is x ∈ D such that a (cid:54) x (cid:54) b in P . Since b ∈ B ( S X ), we have ↓ b ∩ D = X , so x ∈ X .Since a ∈ A ( S X ), there is b ∈ B ( S X ) such that ( a, b ) ∈ S X . It follows that ↓ b ∩ D = X andthus a (cid:54) x (cid:54) b in P , which is a contradiction. (cid:3) The last lemma and Lemma 6.1 imply that the set of incomparable pairs S X is reversible forevery X ∈ X . Since S = S X ∈X S X , it follows that dim( S ) (cid:54) |X | = O ( k ). This completes theproof that dim( S RL ) = O ( k d + k ).7. Connections with graph minors
Recently, a number of important results connecting dimension with structural graph theoryhave been proved [2, 12, 13, 14, 16, 21, 30]. In particular, the following generalization ofTheorem 1.3 is proved in [30] (see [21] for an alternative proof and [13, 16] for further extensions).
Theorem 7.1.
For every positive integer n , there exists an integer d such that if P is a posetthat excludes k and the cover graph of P does not contain K n as a topological minor, then dim( P ) (cid:54) d . Since posets with cover graphs excluding K as a topological minor have dimension boundedindependently of the height [14], Theorem 7.1 is interesting only for n (cid:62)
5. It is natural to askwhether Theorem 1.4 and Conjecture 1.5 can be generalized in the same vein. To address thisquestion, we need to take a closer look on some properties of interval orders.An interval order is a poset P that admits an interval representation , which we define here asan assignment P x ( ‘ x , r x ) ⊂ R of non-empty open intervals to the points of P such that x < y in P if and only if r x (cid:54) ‘ y . Such an interval representation of P is distinguishing if theendpoints ‘ x and r x with x ∈ P are all distinct. It is easy to see that every interval order admitsa distinguishing interval representation. It is well known that P is an interval order if and onlyif P excludes + (the standard example S ) [10].Given a distinguishing interval representation of P , add two new intervals of the form ( a, a + b )and ( a + b , b ) between each pair of consecutive endpoints a and b , and call the resulting intervalorder Q . Then dim( P ) (cid:54) dim( Q ) (as P is a subposet of Q ) and the cover graph of Q has maximumdegree 3. This and the fact that there exist interval orders with arbitrarily large dimension [3]lead to the conclusion (observed by Micek and Wiechert [22]) that there are interval orders witharbitrarily large dimension and with cover graphs of maximum degree 3. Therefore, even for k = 2, Theorem 1.4 and Conjecture 1.5 cannot be generalized to posets that have cover graphsexcluding K or any other graph with maximum degree greater than 3 as a topological minor.By contrast, excluding a minor instead of a topological minor leads to the following observation. Proposition 7.2.
For every positive integer n , there exists an integer d such that if P is aninterval order and the cover graph of P does not contain K n as a minor, then dim( P ) (cid:54) d . Proof.
It is proved in [19] that for every interval order Q , the interval orders that do not contain Q as a subposet have bounded dimension. Fix a positive integer n . We construct a poset Q with ground set { v i : 1 (cid:54) i (cid:54) n } ∪ { e i,j : 1 (cid:54) i < j (cid:54) n } , where the only cover relations are v i < e i,j < v j for 1 (cid:54) i < j (cid:54) n . It is an interval order witnessed by the interval representationthat maps v i to the interval (2 i − , i ) for 1 (cid:54) i (cid:54) n and e i,j to the interval (2 i, j −
1) for1 (cid:54) i < j (cid:54) n . By the result of [19], there is an integer d such that if P is an interval order withdim( P ) > d , then P contains Q as a subposet. It remains to prove that if P contains Q as asubposet, then the cover graph of P contains K n as a minor.Let P be a poset that contains Q as a subposet. For any two points a and b with a < b in P ,let [ a, b ) P = { x ∈ P : a (cid:54) x < b in P } and ( a, b ) P = { x ∈ P : a < x < b in P } . Let V j = j − [ i =1 [ e i,j , v j ) P ∪ { v j } ∪ n [ k = j +1 ( v j , e j,k ) P for 1 (cid:54) j (cid:54) n. It is clear that the subgraphs of the cover graph of P induced on V , . . . , V n are connected. Iteasily follows from the definition of Q that the sets V , . . . , V n are mutually disjoint. A K n minorin the cover graph of P is obtained by contracting the sets V , . . . , V n and deleting the verticesnot in any of V , . . . , V n . Indeed, for 1 (cid:54) i < j (cid:54) n , the cover graph of P contains an edge xe i,j connecting V i and V j , where x is a maximal element in the subset [ v i , e i,j ) P of V i and e i,j ∈ V j . (cid:3) In view of the discussion above, it is natural to make the following conjectures.
Conjecture 7.3.
For every pair ( k, n ) of positive integers, there exists an integer d such thatif P is a poset that excludes k + k and the cover graph of P does not contain K n as a minor,then dim( P ) (cid:54) d . Conjecture 7.4.
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E-mail : [email protected](William T. Trotter) School of Mathematics, Georgia Institute of Technology, Atlanta, GA 30332, USA
E-mail : [email protected](Bartosz Walczak) Department of Theoretical Computer Science, Faculty of Mathematics and Computer Science,Jagiellonian University, Kraków, Poland
E-mail : [email protected](Ruidong Wang) Blizzard Entertainment, Irvine, CA 92618, USA