On the dimension of posets with cover graphs of treewidth $2$

ON THE DIMENSION OF POSETS WITH COVER GRAPHS OF TREEWIDTH 2 GWENAËL JORET, PIOTR MICEK, WILLIAM T. TROTTER, RUIDONG WANG, AND VEIT WIECHERT arXiv:1406.3397v4 [math.CO] 4 May 2016 Abstract. In 1977, Trotter and Moore proved that a poset has dimension at most 3 whenever its cover graph is a forest, or equivalently, has treewidth at most 1. On the other hand, a well-known construction of Kelly shows that there are posets of arbitrarily large dimension whose cover graphs have treewidth 3. In this paper we focus on the boundary case of treewidth 2. It was recently shown that the dimension is bounded if the cover graph is outerplanar (Felsner, Trotter, and Wiechert) or if it has pathwidth 2 (Biró, Keller, and Young). This can be interpreted as evidence that the dimension should be bounded more generally when the cover graph has treewidth 2. We show that it is indeed the case: Every such poset has dimension at most 1276. 1. Introduction The purpose of this paper is to show the following: Theorem 1. Every poset whose cover graph has treewidth at most 2 has dimension at most 1276. Let us provide some context for our theorem. Already in 1977, Trotter and Moore [11] showed that if the cover graph of a poset P is a forest then dim(P ) 6 3 and this is best possible, where dim(P ) denotes the dimension of P . Recalling that forests are exactly the graphs of treewidth at most 1, it is natural to ask how big can the dimension be for larger treewidths. Motivated by this question, we proceed with a brief survey of relevant results about the dimension of posets and properties of their cover graphs. One such result, due to Felsner, Trotter and Wiechert [3], states that if the cover graph of a poset P is outerplanar then dim(P ) 6 4. Again, the bound is best possible. Note that outerplanar graphs have treewidth at most 2. Note also that one cannot hope for a similar bound on the dimension of posets with a planar cover graph. Indeed, already in 1981 Kelly [6] presented a family of posets {Qn }n>2 with planar cover graphs and dim(Qn ) = n (see Figure 1). One interesting feature of Kelly’s construction for our purposes is that the cover graphs also have treewidth at most 3 (with equality for n > 5), as is easily verified. In fact, they even have pathwidth at most 3 (with equality for n > 4). Very recently, Biró, Keller and Young [1] showed that if the cover graph of a poset P has pathwidth at most 2, then its dimension is bounded: it is at most 17. Furthermore, they proved that the treewidth of the cover graph of any poset containing the standard example Sn with n > 5 is at least 3, thus showing in particular that Kelly’s construction cannot be modified to have treewidth 2. To summarize, while the dimension of posets with cover graphs of treewidth 3 is unbounded, no such property is known to hold for the case of treewidth 2, and we cannot hope to obtain it by constructing posets containing large standard examples. Moreover, as mentioned above, the dimension is bounded for two important classes of graphs of treewidth at most 2, outerplanar graphs and graphs of pathwidth at most 2. All this can be interpreted as strong evidence that the dimension should be bounded more generally when the cover graph has treewidth at most 2, which is exactly what we prove in this paper. We note that the bound on the dimension we obtain is large (1276), and is most likely far from the truth. Furthermore, while we strove to make our arguments as simple as possible—and as a result did not try to optimize the bound—the proofs are lengthy and technical. We believe that there is still room for improvements, and it could very well be that a different approach would give a better bound and/or more insight into these problems. We conclude this introduction by briefly mentioning a related line of research. Recently, new bounds for the dimension were found for certain posets of bounded height. Streib and Trotter [8] proved that for Date: August 13, 2018. 2010 Mathematics Subject Classification. 06A07, 05C35. Key words and phrases. Poset, dimension, treewidth. G. Joret was supported by a DECRA Fellowship from the Australian Research Council. P. Micek is supported by the Mobility Plus program from The Polish Ministry of Science and higher Education. V. Wiechert is supported by the Deutsche Forschungsgemeinschaft within the research training group ‘Methods for Discrete Structures’ (GRK 1408). 1 b1 b2 b3 b4 b5 b6 z1 z2 z z3 4 w4 w5 w3 w2 w1 a6 z5 a5 a4 a3 a2 a1 Figure 1. Kelly’s construction of a poset Qn with a planar cover graph containing the standard example Sn as a subposet, for n = 6. (Let us recall that the standard example Sn is the poset on 2n elements consisting of n minimal elements a1 , . . . , an and n maximal elements b1 , . . . , bn which is such that ai < bj in Sn if and only if i 6= j.) The subposet induced by the ai ’s and bi ’s form S6 , which has dimension 6. The general definition of Qn for any n > 2 is easily inferred from the figure. Since the standard example Sn has dimension n, this shows that posets with planar cover graphs have unbounded dimension. every positive integer h, there is a constant c such that if a poset P has height at most h and its cover graph is planar, then dim(P ) 6 c. Joret, Micek, Milans, Trotter, Walczak, and Wang [4] showed that for every positive integers h and t, there is a constant c so that if P has height at most h and the treewidth of its cover graph is at most t, then dim(P ) 6 c. These two results are closely related. In particular, one can deduce the result for planar cover graphs from the result for bounded treewidth cover graphs using a ‘trick’ introduced in [8] that reduces the problem to the special case where there is a special minimal element a0 in the poset that is smaller than all the maximal elements. This implies that the diameter of the cover graph is bounded from above by a function of the height of the poset, and it is well-known that planar graphs with bounded diameter have bounded treewidth (see for instance [2]). This trick of having a special minimal element a0 below all maximal elements turned out to be very useful in the context of this paper as well (though for different reasons), see Observation 6 in Section 2. Finally, we mention that several new results on bounding the dimension of certain posets in terms of their height have recently been obtained [12, 7, 5], the interested reader is referred to [5] for a detailed overview of that area. The paper is organized as follows. In Section 2 we give the necessary definitions and present a number of reductions, culminating in a more technical version of our theorem, Theorem 7. Then, in Section 3, we prove the result. 2. Definitions and Preliminaries Let P = (X, 6) be a finite poset. The cover graph of P , denoted cover(P ), is the graph on the elements of P where two distinct elements x, y are adjacent if and only if they are in a cover relation in P ; that is, either x < y or x > y in P , and this relation cannot be deduced from transitivity. Informally, the cover graph of P can be thought of as its order diagram seen as an undirected graph. The dimension of P , denoted dim(P ), is the least positive integer d for which there are d linear extensions L1 , . . . , Ld of P so that x 6 y in P if and only if x 6 y in Li for each i ∈ {1, . . . , d}. We mention that an introduction to the theory of posets and their dimension can be found in the monograph [9] and in the survey article [10]. When x and y are distinct elements in P , we write x k y to denote that x and y are incomparable. Also, we let Inc(P ) = {(x, y) | x, y ∈ X and x k y in P } denote the set of ordered pairs of incomparable elements in P . We denote by min(P ) the set of minimal elements in P and by max(P ) the set of 2 maximal elements in P . The downset of a set S ⊆ X of elements is defined as D(S) = {x ∈ X | ∃s ∈ S such that x 6 s in P }, and similarly we define the upset of S to be U (S) = {x ∈ X | ∃s ∈ S such that s 6 x in P }. A set I ⊆ Inc(P ) of incomparable pairs is reversible if there is a linear extension L of P with x > y in L for every (x, y) ∈ I. It is easily seen that if P is not a chain, then dim(P ) is the least positive integer d for which there exists a partition of Inc(P ) into d reversible sets. A subset {(xi , yi )}ki=1 of Inc(P ) with k > 2 is said to be an alternating cycle if xi 6 yi+1 in P for each i ∈ {1, 2, . . . , k}, where indices are taken cyclically (thus xk 6 y1 in P is required). For example, in the poset Q6 of Figure 1 the pairs (ai , bi ), (aj , bj ) form an alternating cycle of length 2 for all i, j ∈ {1, . . . , 6} such that i 6= j. An alternating cycle {(xi , yi )}ki=1 is strict if, for each i, j ∈ {1, 2, . . . , k}, we have xi 6 yj in P if and only if j = i + 1 (cyclically). Note that in that case x1 , x2 , . . . , xk are all distinct, and y1 , y2 , . . . , yk are all distinct. Notice also that every non-strict alternating cycle can be made strict by discarding some of its incomparable pairs. Observe that if I = {(xi , yi )}ki=1 is an alternating cycle in Inc(P ) then I cannot be reversed by a linear extension L of P . Indeed, otherwise we would have yi < xi 6 yi+1 in L for each i ∈ {1, 2, . . . , k}, which cannot hold cyclically. Hence, alternating cycles are not reversible. It is easily checked—and this was originally observed by Trotter and Moore [11]—that every non-reversible subset I ⊆ Inc(P ) contains an alternating cycle, and thus a strict alternating cycle: Observation 2. A set I of incomparable pairs of a poset P is reversible if and only if I contains no strict alternating cycle. An incomparable pair (x, y) of a poset P is said to be a min-max pair if x is minimal in P and y is maximal in P . The set of all min-max pairs in P is denoted by MM(P ). Define dim∗ (P ) as the least positive integer t such that MM(P ) can be partitioned into t reversible subsets if MM(P ) 6= ∅, and as being equal to 1 otherwise. For our purposes, when bounding the dimension we will be able to focus on reversing only those incomparable pairs that are min-max pairs. This is the content of Observation 3 below. In order to state this observation formally we first need to recall some standard definitions from graph theory. By ‘graph’ we will always mean an undirected finite simple graph in this paper. The treewidth of a graph G = (V, E) is the least positive integer t such that there exist a tree T and non-empty subtrees Tx of T for each x ∈ V such that (i) V (Tx ) ∩ V (Ty ) 6= ∅ for each edge xy ∈ E, and (ii) |{x ∈ V | u ∈ V (Tx )}| 6 t + 1 for each node u of the tree T . The pathwidth of G is defined as treewidth, except that the tree T is required to be a path. A graph H is a minor of a graph G if H can be obtained from a subgraph of G by contracting edges. (We note that since we only consider simple graphs, loops and parallel edges resulting from edge contractions are deleted.) Recall that the class of graphs of treewidth at most k (k > 0) is closed under taking minors, thus tw(H) 6 tw(G) for every graph G and minor H of G. Given a class F of graphs, we let Fb denote the class of graphs that can be obtained from a graph G ∈ F by adding independently for each vertex v of G zero, one, or two new pendant vertices adjacent to v. We will use the easy observation that Fb = F when F is the class of graphs of treewidth at most k (provided k > 1). We note that Fb = F holds for other classes F of interest, such as planar graphs. The next elementary observation is due to Streib and Trotter [8], who were interested in the case of planar cover graphs. We provide a proof for the sake of completeness. Observation 3. Let F be a class of graphs. If P is a poset with cover(P ) ∈ F then there exists a poset Q such that b and (i) cover(Q) ∈ F, (ii) dim(P ) 6 dim∗ (Q). Proof. If P is a chain then we set Q = P and the statement can be easily verified. Otherwise, let Q be the poset constructed from P as follows: For each non-minimal element x of P , add a new element x′ below x (and its upset) such that x′ < x is the only cover relation involving x′ in Q. Also, for each non-maximal element y of P , add a new element y ′′ above y (and its downset) such that y < y ′′ is the only cover relation involving y ′′ in Q. Now, the cover graph of Q is the same as the cover graph of P except that we attached up to two new pendant vertices to each vertex. For convenience, we also define an element x′ for each minimal element x of P , simply by setting ′ x = x. Similarly, we let y ′′ = y, for each maximal element y of P . 3 Observe that if a set L of linear extensions of Q reverses all min-max pairs of Q then it must reverse all incomparable pairs of P . Indeed, for each pair (x, y) ∈ Inc(P ) consider the min-max pair (x′ , y ′′ ) in Q. There is some linear extension L ∈ L reversing (x′ , y ′′ ). Given that x′ 6 x and y 6 y ′′ in Q, it follows that y 6 y ′′ < x′ 6 x in L. Hence, restricting the linear orders in L to the elements of P we deduce that L reverses all pairs in Inc(P ) so dim(P ) 6 |L| (as P is not a chain). Therefore, dim(P ) 6 dim∗ (Q).  As a corollary, for treewidth we obtain: Observation 4. For every poset P there exists a poset Q such that (i) tw(cover(P )) = tw(cover(Q)), and (ii) dim(P ) 6 dim∗ (Q). Proof. This follows from Observation 3 if tw(cover(P )) > 1. If, on the other hand, tw(cover(P )) = 0, then P is an antichain and we can simply take Q = P .  In the next observation we consider posets with disconnected cover graphs. As expected, we define the components of a poset P as the subposets of P induced by the components of its cover graph. Observation 5. If P is a poset with k > 2 components C1 , . . . , Ck then either (i) P is a disjoint union of chains and we have dim(P ) = dim∗ (P ) = 2, or (ii) dim(P ) = max{dim(Ci ) | i = 1, . . . , k} and dim∗ (P ) = max{dim∗ (Ci ) | i = 1, . . . , k}. Proof. If for each i ∈ {1, . . . , k} the subposet Ci of P is a chain then it is easy to see that dim(P ) = dim∗ (P ) = 2. Thus we may assume that this is not the case, that is, dim(Ci ) > 2 for some i ∈ {1, . . . , k}. For each i ∈ {1, . . . , k} let Ri be a family of dim(Ci ) linear extensions of Ci witnessing the dimension of Ci . We construct a family R of linear extensions of P in the following way. First, let R := ∅. Then, as long as there is a set Ri which is not empty, (i) choose a linear extension Li ∈ Ri for each i ∈ {1, . . . , k} such that Ri is not empty; (ii) choose any linear extension Li of Ci for each i ∈ {1, . . . , k} such that Ri is empty; (iii) add to R the linear extension L of P defined by L := L1 < L2 < . . . < Lk , and (iv) remove Li from Ri for each i ∈ {1, . . . , k}. Clearly, |R| = max{dim(Ci ) | i ∈ {1, . . . , k}}. Now consider one arbitrarily chosen linear extension L ∈ R; say we had L = L1 < . . . < Lk when it was defined above, and replace L by L′ := Lk < . . . < L1 in R. It is easy to verify that the resulting family R reverses all incomparable pairs in P . In particular, all incomparable pairs of P with elements from distinct components are reversed by L′ and any other linear extension in R (note there is at least one more as dim(Ci ) > 2 for some i). This shows that dim(P ) = max{dim(Ci ) | i ∈ {1, . . . , k}}. The proof for dim∗ (P ) goes along the same lines and is thus omitted.  To prove the next observation we partition the minimal and maximal elements of a poset by ‘unfolding’ the poset from an arbitrary minimal element, and contract some part of the poset into a single element. This proof idea is due to Streib and Trotter [8], and is very useful for our purposes. In [8] it was used in the context of planar cover graphs but it works equally well for any minor-closed class of graphs. Observation 6. For every poset P there exists a poset Q such that (i) cover(Q) is a minor of cover(P ) (and thus in particular tw(cover(Q)) 6 tw(cover(P ))); (ii) there is an element q0 ∈ min(Q) with q0 < q in Q for all q ∈ max(Q), and (iii) dim∗ (P ) 6 2 dim∗ (Q). Proof. First of all, we note that it is enough to prove the statement in the case where cover(P ) is connected. Indeed, if cover(P ) is disconnected then by Observation 5 either P is a disjoint union of chains and dim(P ) = dim∗ (P ) = 2, in which case the observation is trivial, or dim∗ (P ) = max{dim∗ (C) | C component of P } and we can simply consider a component C of P with dim∗ (P ) = dim∗ (C). From now on we suppose that cover(P ) is connected. We are going to build a small set of linear extensions of P reversing all min-max pairs of P . Partition the minimal and maximal elements of P as follows. Choose an arbitrary element a0 ∈ min(P ), let A0 = {a0 }, and for i = 1, 2, 3, . . . let [ Bi = {b ∈ max(P ) − Bj | there exists a ∈ Ai−1 with a < b in P }, 16j<i Ai = {a ∈ min(P ) − [ Aj | there exists b ∈ Bi with a < b in P }. 06j<i 4 Bk−1 B2 B1 Bk Ak−1 A2 A1 a0 Figure 2. Schematic drawing of P and the sets A0 , A1 , . . . , Ak−1 and B1 , . . . , Bk . Bi Bi+1 Ai Q Qi+1 i Bi+1 q Ai+1 Ai Figure 3. Definition of Qii+1 and construction of Q with its cover graph. Let k be the least index such that Ak is empty. See Figure 2 for an illustration. The fact that each minimal and maximal element of P is included in one of the sets defined above follows from the connectivity of cover(P ). If k = 1 then a0 is below all maximal elements of P and hence P itself satisfies conditions (i)-(iii). So we may assume k > 2 from now on.
Let Qi+1 be the poset on the set of elements Xii+1 = Ai ∪ Bi+1 ∪ U (Ai ) ∩ D(Bi+1 ) with order i relation inherited from P . Figure 3 illustrates this definition. Let t = max{dim∗ (Qi+1 ) | i = 0, . . . , k − 1}. i For each i ∈ {0, . . . , k − 1} consider t linear extensions Li1 , . . . , Lit of Qi+1 that reverse all pairs from the i set MM(Qi+1 ). Combining these we define t linear extensions of P . For j ∈ {1, . . . , t} let Lj be a linear i extension of P that contains the linear order Ljk−1 < · · · < L1j < L0j . Then, L1 , . . . , Lt reverse all pairs (a, b) ∈ MM(P ) with a ∈ Ai and b ∈ Bj where j > i + 1. In a similar way we are able to reverse the pairs where j 6 i.
Let Qii be the poset on the set of elements Xii = Ai ∪ Bi ∪ U (Ai ) ∩ D(Bi ) being ordered as in P . We set t′ = max{dim∗ (Qii ) | i ∈ {1, . . . , k − 1}} and for each i ∈ {1, . . . , k − 1} we fix t′ linear extensions Li1 , . . . , Lit′ of Qii reversing all pairs from MM(Qii ). Again, we combine these to obtain linear extensions of P . For j ∈ {1, . . . , t′ } let L′j be a linear extension of P that contains the linear order L1j < L2j < · · · < Lk−1 . j Clearly, L′1 . . . , L′t′ reverse all pairs (a, b) ∈ MM(P ) with a ∈ Ai and b ∈ Bj where j 6 i. It follows that L1 , . . . , Lt , L′1 . . . , L′t′ reverse the set MM(P ) and hence dim∗ (P ) 6 t + t′ . Now suppose first t > t′ , so in particular t > 1. Then let ℓ ∈ {0, . . . , k − 1} such that t = dim∗ (Qℓ+1 ). ℓ Note that we must have ℓ > 1 since dim∗ (Q10 ) = 1 < t. We define Q to be the poset that is obtained by adding an extra element q which is such that q > x for all x ∈ Xℓℓ+1 ∩ D(Bℓ ), and from Qℓ+1 ℓ incomparable to all other elements of Qℓ+1 (here we need ℓ > 1 so that Bℓ exists). In particular, q > a ℓ for all a ∈ Aℓ . Observe that the cover graph of Q is an induced subgraph of cover(P ) with an extra vertex q linkedSto some of the other vertices. Here, q can be seen as the result of the contraction of the connected set 16j6ℓ D(Bj ) − Xℓℓ+1 plus the deletion of some of the edges incident to the contracted vertex (see Figure 3 with i = ℓ, dashed edges indicate deletions). The deletion step is necessary, as after the contraction it might be that some edges incident to q do not correspond to cover relations anymore. It follows that cover(Q) is a minor of cover(P ). Furthermore, it holds that ) 6 2 dim∗ (Q). dim∗ (P ) 6 2t = 2 dim∗ (Qℓ+1 ℓ Therefore, the dual of Q satisfies conditions (i)-(iii). The case t′ > t goes along similar lines as in the first case (with the slight difference that we do not need to exclude the subcase t′ = 1). We leave the details to the reader.  5 Applying these observations we move from Theorem 1 to a more technical statement. Theorem 7. Let P be a poset with (i) a cover graph of treewidth at most 2, and (ii) a minimal element a0 ∈ min(P ) such that a0 < b for all b ∈ max(P ). Then the set MM(P ) can be partitioned into 638 reversible sets. In order to deduce Theorem 1 from Theorem 7 consider any poset P with cover graph of treewidth at most 2. By Observation 4 there is a poset Q with tw(cover(Q)) 6 2 and dim(P ) 6 dim∗ (Q). Now by Observation 6 and applying Theorem 7 there is a poset R with tw(cover(R)) 6 2, a minimal element a0 ∈ min(R) such that a0 < b for all b ∈ max(R), and dim(P ) 6 dim∗ (Q) 6 2 dim∗ (R) 6 2 · 638 = 1276, as desired. From now on we focus on the proof of Theorem 7. Let P = (X, 6) be a poset fulfilling the conditions of Theorem 7. Consider a tree decomposition of width at most 2 of cover(P ), consisting of a tree T and subtrees Tx for each x ∈ X. We may assume that the width of the decomposition is exactly 2, since otherwise dim(P ) 6 3 by the result of Trotter and Moore [11], and the theorem follows trivially. For each node u of T let B(u) denote its bag, namely, the set {x ∈ X | u ∈ V (Tx )}. Since the tree decomposition has width 2, every bag has size at most 3, and at least one bag has size exactly 3. Modifying the tree decomposition if necessary, we may suppose that every bag has size 3. Indeed, say uv is an edge of T with |B(u)| = 3 and |B(v)| 6 2. Then choose arbitrarily 3 − |B(v)| elements from B(u) \ B(v) and add them to B(v). Repeating this process as many times as necessary, we eventually ensure that every bag has size 3. Note that the subtrees Tx (x ∈ X) of the tree decomposition are uniquely determined by the bags, and vice versa; thus, it is enough to specify how T and the bags are modified. The above modification repeatedly adds leaves to some of the subtrees Tx (x ∈ X), which clearly keeps the fact that T and the subtrees Tx (x ∈ X) form a tree decomposition of cover(P ). Recall that, by the assumptions of Theorem 7, the poset P has a minimal element a0 with a0 < b for all b ∈ max(P ). This implies that the cover graph of P is connected. Using this, we may suppose without loss of generality that |B(u) ∩ B(v)| > 1 for each edge uv of T . For if this does not hold, then the bags of one of the two components of T − uv are all empty (as is easily checked), and thus the nodes of that component can be removed from T without affecting the tree decomposition. In fact, we may even assume that |B(u) ∩ B(v)| = 2 holds for every edge uv of T . To see this, consider the following iterative modification of the tree decomposition: Suppose that uv is an edge of T such that t := |B(u) ∩ B(v)| 6= 2. If t = 3 then simply identify u and v, and contract the edge uv in T . If t = 1 then subdivide the edge uv in T with a new node w, and let the bag B(w) of w be the set (B(u) ∩ B(v)) ∪ {x, y}, where x and y are arbitrarily chosen elements in B(u) \ B(v) and B(v) \ B(u), respectively. These modifications are valid, in the sense that the bags still define a tree decomposition of cover(P ) of width 2, and in order to ensure the desired property it suffices to apply them iteratively until there is no problematic edge left. To summarize, in the tree decomposition we have |B(u)| = 3 for every node u of T , and |B(u)∩B(v)| = 2 for every edge uv of T . We will need to further refine our tree decomposition so as to ensure a few extra properties. These changes will be explained one by one below. Let us mention that we will keep the fact that |B(u) ∩ B(v)| = 2 for every edge uv of T , and that |B(u)| = 3 for every internal node u of T . However, we will add new leaves to T having bags of size 2 only. Choose an arbitrary node r′ ∈ V (T ) with a0 ∈ B(r′ ). Add a new node r to T and make it adjacent to ′ r . The bag B(r) of r is defined as the union of a0 and one arbitrarily chosen element from B(r′ ) − {a0 }. (Observe that the size of B(r) is only 2; on the other hand, we do have |B(r) ∩ B(r′ )| = 2.) We call r the root of T , and thus see T as being rooted at r. (For a technical reason we need the root to be a leaf of T , which explains why we set it up this way.) Every non-root node u in T has a parent p(u) in T , namely, the neighbor of u on the path from u to r in T . Now we have an order relation on the nodes of T , namely u 6 v in T if u is on the path from r to v in T . The following observation will be useful later. Observation 8. If v1 , . . . , vn is a sequence of nodes of T such that consecutive nodes are comparable in T (that is vi 6 vi+1 or vi+1 6 vi in T for each i ∈ {1, . . . , n − 1}), then there is an index j ∈ {1, . . . , n} such that vj 6 vi in T for each i ∈ {1, . . . , n}. Proof. We prove this by induction on n. For n = 1 it is immediate. So suppose that n > 1. Then we can apply the induction hypothesis on the sequence v1 , . . . , vn−1 and get j ∈ {1, . . . , n − 1} such that vj 6 vi 6 for each i ∈ {1, . . . , n − 1}. As vn−1 and vn are comparable in T , we have vn−1 6 vn or vn 6 vn−1 in T . In the first case we conclude vj 6 vn−1 6 vn in T and we are done. In the second case we have {vj , vn } 6 vn−1 in T , which makes vj and vn comparable in T . But clearly, from this it follows that vj 6 vi in T for each i ∈ {1, . . . , n} or vn 6 vi in T for each i ∈ {1, . . . , n}.  Fix a planar drawing of the tree T with the root r at the bottom. Suppose that v and v ′ are two nodes of T that are incomparable in T . Take the maximum node u (with respect to the order in T ) such that u 6 v and u 6 v ′ in T . We denote this node by v ∧ v ′ . Observe that u has degree at least 2 in T , and hence is distinct from the root r. (Ensuring this is the reason why we made sure that the root r is a leaf.) Consider the edge p from u to p(u), the edge e from u towards v and the edge e′ from u towards v ′ . All these edges are distinct. If the clockwise order around u in the drawing is p, e, e′ for these three edges, then we say that v is to the left of v ′ in T , otherwise the clockwise order around u is p, e′ , e and we say that v is to the right of v ′ in T . Observe that the relations “is left of in T ” and “is right of in T ” both induce a linear order on any set of nodes which are pairwise incomparable in T . Observation 9. Let v and v ′ be incomparable nodes in T with v left of v ′ in T , and let u := v ∧ v ′ . If w and w′ are the neighbors of u on the paths towards v and v ′ in T , respectively, then for each node c in T we have that (i) v is left of c in T if w′ 6 c in T , and (ii) c is left of v ′ in T if w 6 c in T . Proof. If w′ 6 c in T , then we also have u = v ∧ c, and the first edge on the path from u to c in T is the same as that of the path from u to v ′ in T . Since v is left of v ′ in T , it follows that v is left of c as well. The proof for the second item is analogous.  Next we modify once more the tree decomposition. For each element a ∈ min(X) such that (a, b) ∈ MM(P ) for some b ∈ max(X), choose arbitrarily a node wa of T such that a ∈ B(wa ). Similarly, for each element b ∈ max(X) such that (a, b) ∈ MM(P ) for some a ∈ min(X), choose arbitrarily a node wb of T such that b ∈ B(wb ). (Note that the same node of T could possibly be chosen more than once.) Now that all these choices are made, for each minimal element a of P considered above, add a new leaf aT to T adjacent to wa with bag B(aT ) := {a, x}, where x is an arbitrarily chosen element from B(wa ) \ {a}. Similarly, for each maximal element b of P considered above, add a new leaf bT to T adjacent to wb with bag B(bT ) := {b, x}, where x is an arbitrarily chosen element from B(wb ) \ {b}. This concludes our modifications of the tree decomposition. Notice that we made sure that |B(u)| = 3 for every internal node u of T , and that |B(u) ∩ B(v)| = 2 for every edge uv of T . Observe also that for every pair (a, b) ∈ MM(P ), the two nodes aT and bT are incomparable in T , and thus one is to the left of the other in T . Figure 4 provides an illustration. (We also note that while the tree T has been modified since stating Observations 8 and 9, they obviously still apply to the new tree T .) Let G be the intersection graph of the subtrees Tx (x ∈ X) of T . Thus two distinct elements x, y ∈ X are adjacent in G if and only if V (Tx ) ∩ V (Ty ) 6= ∅. The graph G is chordal and the maximum clique size in G is 3. Hence the vertices of G can be (properly) colored with three colors. We fix a 3-coloring φ of X which is such that x, y ∈ X receive distinct colors whenever V (Tx ) ∩ V (Ty ) 6= ∅. In particular, if x and y are two distinct elements of P such that x, y ∈ B(u) for some u ∈ V (T ) then x and y receive different colors. We end this section with a fundamental observation which is going to be used repeatedly in a number of forthcoming arguments. We say that a relation x 6 y in P hits a set Z ⊆ X if there exists z ∈ Z with x 6 z 6 y in P . Observation 10. Let x 6 y in P and let u, v ∈ V (T ) be such that x ∈ B(u), y ∈ B(v). (i) If w ∈ V (T ) lies on the path from u to v in T then x 6 y hits B(w). (ii) If e = w1 w2 ∈ E(T ) lies on the path from u to v in T then x 6 y hits B(w1 ) ∩ B(w2 ). (iii) If w1 , . . . , wt ∈ V (T ) are t nodes on the path from u to v in T appearing in this order, then there exist zi ∈ B(wi ) for each i ∈ {1, . . . , t} such that x 6 z1 6 · · · 6 zt 6 y in P . Proof. Suppose that w lies on a path from u to v in T . Since x 6 y in P there is a path x = z0 , zS 1 , . . . , zk = y in G such that zi < zi+1 is a cover relation in P for each i ∈ {0, 1, . . . , k−1}. This means that 06i6k Tzi S is a (connected) subtree of T containing u and v. Thus, 06i6k Tzi contains w and therefore there exists i with zi ∈ B(w). The proof of (ii) is analogous. We prove (iii) by induction on t. For t = 1 this corresponds to (i), so let us assume t > 1 and consider the inductive case. By induction there exist zi ∈ B(wi ) for each i ∈ {1, . . . , t − 1} such that 7 aT bT Ta Tb u Ta0 p(u) r Figure 4. We have the following properties in this example: (a, b) ∈ MM(P ) with aT left of bT in T , and u = aT ∧ bT (and hence u < aT and u < bT in T ). We also have B(u) = {a0 , a, b} here. ν10 ν15 2 2 ν14 ν9 4 2 ν7 ν8 no 6 yes ν6 ν13 ν12 6 6 ν11 ν5 no ν3 no ν4 yes yes ν2 2 ν1 Figure 5. The signature tree Ψ. Each of the three branching nodes ν2 , ν4 , and ν6 corresponds to a yes/no question, and the edges towards their children are labeled according to the possible answers. Edges starting from a non-branching node νi to a node νj are labeled with the size of Σ(νi , νj ). x 6 z1 6 · · · 6 zt−1 6 y in P . Applying (i) with relation zt−1 6 y and the wt−1 –v path, we obtain that zt−1 6 zt 6 y in P for some zt ∈ B(wt ). Combining, we obtain x 6 z1 6 · · · 6 zt 6 y in P , as desired.  3. The Proof We aim to partition MM(P ) into a constant number of sets, each of which is reversible. This will be realized with the help of a signature tree, which is depicted on Figure 5. This plane tree Ψ, rooted at node ν1 , assigns to each pair (a, b) ∈ MM(P ) a corresponding leaf of Ψ according to properties of the pair (a, b). 8 The nodes ν1 , . . . , ν15 of Ψ are enumerated by depth-first and left-to-right search. Each node νi which is distinct from the root and not a leaf has a corresponding function of the form αi : MM(P, νi ) → Σi , where MM(P, νi ) ⊆ MM(P ) and Σi is a finite set, whose size does not depend on P . We put MM(P, ν1 ) = MM(P ) and the other domains will be defined one by one in this section. To give an example, let us look forward to upcoming subsections where we define α1 and α2 as follows. • α1 (a, b) ∈ Σ1 = {left, right} encodes whether aT is to the left or to the right of bT in T ; • α2 (a, b) ∈ Σ2 = {yes, no} is the answer to the question “Is there an element q ∈ B(aT ∧ bT ) with a 6 q in P ?”. Furthermore, for each internal node νi with children νi1 , . . . , νil in Ψ, the edges νi νi1 , . . . , νi νil of Ψ are respectively labeled by subsets Σ(νi , νi1 ), . . . , Σ(νi , νil ) of Σi such that Σi = Σ(νi , νi1 ) ⊔ · · · ⊔ Σ(νi , νil ), that is, so that the sets Σ(νi , νij ) form a partition of Σi . For example, Σ(ν1 , ν2 ) = {left, right} = Σ1 ; Σ(ν2 , ν3 ) = {no}; Σ(ν2 , ν4 ) = {yes}. Observe that each internal node νi of Ψ has either one or two children; in particular, if νi has only one child then the corresponding edge is labeled with the full set Σi . The reader may wonder why we do not refine the tree Ψ and have an edge out of νi for every possible value in Σi . This is because sometimes several values in Σi will correspond to analogous cases in our proofs which can be treated all at once. To give a concrete example, consider Σ(ν1 , ν2 ) = {left, right}: When proving that a set S of min-max pairs is reversible, the case that aT is left of bT for every (a, b) ∈ S is analogous to the case that aT is right of bT for every (a, b) ∈ S, as one is obtained from the other by exchanging the notion of left and right in T (that is, by replacing the plane tree T by its mirror image). Hence it will be enough to only consider, say, the case where aT is to the left of bT for every (a, b) ∈ S. Now for an internal node νi of Ψ distinct from the root (i 6= 1), let ν1 = νi1 , . . . , νil = νi be the path from the root ν1 to νi in Ψ. Define the signature of νi as the set Σ(νi ) = Σ(νi1 , νi2 ) × . . . × Σ(νil−1 , νil ) (1) and let MM(P, νi ) = {(a, b) ∈ MM(P ) | (αi1 (a, b), . . . , αil−1 (a, b)) ∈ Σ(νi )}; MM(P, νi , Σ) = {(a, b) ∈ MM(P ) | (αi1 (a, b), . . . , αil−1 (a, b)) = Σ} for Σ ∈ Σ(νi ). Observe that by this definition, for each internal node νi of Ψ with children νi1 , . . . , νil we get the partition [ MM(P, νi ) = MM(P, νij ). 16j6l Therefore, by construction the sets MM(P, νi ) with νi a leaf of Ψ (so for ν3 , ν7 , ν10 , ν15 ) form a partition of MM(P ). With a further refinement it follows that [ [ MM(P ) = MM(P, νi , Σ) νi leaf of Ψ Σ∈Σ(νi ) and the proof below boils down to showing that MM(P, νi , Σ) is reversible for each leaf νi of Ψ and each Σ ∈ Σ(νi ). Once this is established we get an upper bound on dim∗ (P ) just by counting the number of sets in our partition of MM(P ), namely X |Σ(νi )| = |Σ(ν3 )| + |Σ(ν7 )| + |Σ(ν10 )| + |Σ(ν15 )| dim∗ (P ) 6 νi leaf of Ψ = 2 + 2 · 6 + 2 · 6 · 2 · 2 + 2 · 6 · 6 · 4 · 2 = 638. Note that the calculation for the particular summands follows from (1) and the edge labelings in Figure 5. Our proof will follow a depth-first, left-to-right search of the signature tree Ψ, defining the functions αi one by one in that order, and showing that for each Σ ∈ Σ(νi ) the set MM(P, νi , Σ) is reversible when encountering a leaf νi . Hence, the tree Ψ also serves as a road map of the proof. Moreover, for the reader’s convenience we included a table collecting all functions αi and their meanings, 9 Function Section Meaning φ 2 α1 3.1 α2 3.1 α4 3.3 α5 3.3 Proper 3-coloring of the intersection graph of the subtrees Tx (x ∈ X) of T . That is, φ(x) 6= φ(y) whenever V (Tx ) ∩ V (Ty ) 6= ∅. Assigns ‘left’ or ‘right’ to pairs (a, b) depending on whether aT lies to the left or to the right of bT in T . Records the answer to the question: “Is there an element q ∈ B(uab ) such that a 6 q in P ?” Records the answer to the question: “Is there an element q ∈ B(uab ) ∩ B(pab ) such that a 6 q in P ?” α5 (a, b) = (φ(xab ), φ(yab ), φ(zab )), where B(uab ) = {xab , yab , zab } and xab , yab , zab satisfy: • a 6 xab 66 b in P ; • B(uab ) ∩ B(pab ) = {yab , zab }; • a 66 yab 6 b in P ; • a0 6 yab in P ; • a 66 zab in P , and • yab ∈ B(uab ) ∩ B(wab ). α6 3.4 α8 3.6 α9 3.7 α11 3.9 α12 3.10 α13 α14 Records the answer to the question “Is B(uab ) ∩ B(wab ) = {xab , yab }?”. Given (a, b) ∈ MM(P, ν8 ), let Σ ∈ Σ8 be such that (a, b) ∈ MM(P, ν8 , Σ). Then α8 (a, b) = ψ8,Σ (a, b), where ψ8,Σ is a 2-coloring of the graph SΣ of special 2-cycles. Given (a, b) ∈ MM(P, ν9 , Σ) for some Σ ∈ Σ(ν9 ), function α9 records the color ψ9,Σ (a, b), where ψ9,Σ is a coloring of KΣ . α11 (a, b) = (φ(xab ), φ(yab ), φ(zab )), where B(uab ) = {xab , yab , zab } and xab , yab , zab satisfy: • B(uab ) ∩ B(pab ) = {xab , yab }; • a 6 xab 66 yab in P , and • a 66 yab 6 b in P . Given (a, b) ∈ MM(P, ν12 ), α12 (a, b) records the answers to the questions “Is a 6 zab in P ”, “Is zab 6 b in P ?”, and “Is a0 6 xab in P ?”. 3.11 Given (a, b) ∈ MM(P, ν13 , Σ), function α13 records the color of the pair (a, b) in the 4-coloring ψ13,Σ of the graph JΣ . The purpose of α13 is to get rid of 2-cycles in MM(P, ν13 , Σ). 3.12 Given (a, b) ∈ MM(P, ν14 , Σ), function α14 records the color of the pair (a, b) in the 2-coloring ψ14,Σ of the graph K̂Σ . The purpose of α14 is to get rid of strict alternating cycles of length at least 3 in MM(P, ν14 , Σ). Table 1. Table of functions and their meanings see Table 1. It is not necessary to read this table now, but it might be helpful while going through the main proof. Now that the necessary definitions are introduced and the preliminary observations are made, we are about to consider the nodes of the signature tree one by one, stating and proving many technical statements along the way. At this point the reader might legitimately wonder why it all works, that is, 10 bT aT wab vab uab pab r Figure 6. Pair (a, b) ∈ MM(P ) with α1 (a, b) = left and the corresponding nodes pab , uab , vab , and wab in T . what are the basic ideas underlying our approach. While we are unable to offer a general intuition— indeed, this is why we believe that better insights into these posets remain to be obtained—we can at least explain a couple of the strategies we repeatedly apply in our proofs. A first strategy builds on the fact that when choosing three times an element in a 2-element set, some element is bound to be chosen at least twice: As a toy example, suppose that {(ai , bi )}ki=1 is a strict alternating cycle with k > 3 in some subset I ⊆ MM(P ) which we are trying to prove is reversible. Suppose further that we somehow previously established that the aTi –bTi+1 path in T includes a specific edge uv of T for at least three distinct indices i ∈ {1, . . . , k}; which indices is not important, so let us say this happens for indices 1, 2, 3. Then by Observation 10 the relation ai 6 bi+1 hits B(u) ∩ B(v) for each i = 1, 2, 3. Given that |B(u) ∩ B(v)| = 2, this implies that some element x ∈ B(u) ∩ B(v) is hit by two of these relations, that is, we have ai 6 x 6 bi+1 and aj 6 x 6 bj+1 in P for some i, j ∈ {1, 2, 3} with i < j. However, this implies ai 6 x 6 bj+1 in P , contradicting the fact that the alternating cycle is strict. (Here we use that k > 3.) Therefore, the alternating cycle {(ai , bi )}ki=1 could not have existed in the first place. More generally, when analyzing certain situations we claim cannot occur, we will typically easily find two relations c1 6 d1 and c2 6 d2 in P both hitting B(u) ∩ B(v) for some edge uv of T , and which are incompatible, in the sense that they cannot hit the same element. The work then goes into pinning down a third relation c3 6 d3 in P which is incompatible with the first two, and yet hits B(u) ∩ B(v). (The fact that a0 6 b in P for every b ∈ max(P ) will often be helpful here.) A second strategy is to see certain strict alternating cycles as inducing a graph on MM(P ), and then study and exploit properties of said graph. This is natural for strict alternating cycles of length 2: Any such cycle (a1 , b1 ), (a2 , b2 ) can be seen as inducing an edge between vertex (a1 , b1 ) and vertex (a2 , b2 ). If we somehow can show that the resulting graph has bounded chromatic number, then we can consider a corresponding coloring of the pairs, and we will know that within a color class there are no strict alternating cycles of length 2 left. Thus, by doing so we ‘killed’ all such cycles by partitioning the pairs in a constant number of sets. Such a strategy is used twice in the proof, when considering nodes ν8 and ν13 of the signature tree Ψ. We also use a variant of it tailored to handle certain strict alternating cycles of length at least 3 and involving a directed graph on MM(P ), when considering nodes ν9 and ν14 of Ψ. We now turn to the proof. From now on we will use the following notations for a given pair (a, b) ∈ MM(P ): We let uab := aT ∧ bT , pab := p(uab ), and denote by vab and wab the neighbors of uab in T towards aT and bT , respectively. Figure 6 illustrates the newly defined nodes. 3.1. Nodes ν1 and ν2 and their respective functions α1 and α2 . We start with the definition of α1 , which belongs to the node ν1 of the signature tree Ψ. For each (a, b) ∈ MM(P, ν1 ) = MM(P ), α1 (a, b) ∈ Σ1 = {left, right} encodes whether aT is to the left or to the right of bT in T . Let us proceed with node ν2 and its function α2 . 11 For each (a, b) ∈ MM(P, ν2 ) = MM(P ), we let α2 (a, b) ∈ Σ2 = {yes, no} be the answer to the question “Is there an element q ∈ B(uab ) with a 6 q in P ?”. Note that it might be the case that all elements of B(uab ) are incomparable with a (so the answer is “no”), while there is always an element q ∈ B(uab ) such that q 6 b in P as the comparability a0 6 b hits the bag B(uab ). In the next section we treat all the pairs (a, b) ∈ MM(P ) with α2 (a, b) = no and it turns out that they can easily be reversed. 3.2. First leaf of Ψ: the node ν3 . Incomparable pairs of MM(P, ν3 ) received one of the two possible signatures in Σ(ν3 ) = {(left, no), (right, no)}. We start by showing that pairs with these signatures are reversible. Claim 11. MM(P, ν3 , Σ) is reversible for each Σ ∈ Σ(ν3 ). Proof. Let Σ ∈ Σ(ν3 ) = {(left, no), (right, no)}. We will assume that Σ = (left, no), thus α1 (a, b) = left for pairs (a, b) ∈ MM(P, ν3 , Σ). In the other case it suffices to exchange the notion of left and right in the following argument. (We note that we will start with that assumption in all subsequent proofs, for the same reason.) Arguing by contradiction, suppose that there is a strict alternating cycle {(ai , bi )}ki=1 in MM(P, ν3 , Σ). Thus ai 6 bi+1 in P for all i (cyclically). Let bTj be leftmost in T among all the bTi ’s (i ∈ {1, . . . , k}). The node aTj is to the left of bTj (as (aj , bj ) ∈ MM(P, ν3 , Σ), so α1 (aj , bj ) = left), and thus to the left of all the bTi ’s. Hence, the path from aTj to bTj+1 in T goes through the node uaj bj . By Observation 10, the relation aj 6 bj+1 in P hits B(uaj bj ), contradicting α2 (aj , bj ) = no (recall that (aj , bj ) ∈ MM(P, ν3 , Σ), and thus α2 (aj , bj ) = no).  As a consequence of Claim 11, all the pairs (a, b) ∈ MM(P ) being considered in the following satisfy α2 (a, b) = yes (see how this fact can be read from the signature tree Ψ). 3.3. Nodes ν4 and ν5 and their respective functions α4 and α5 . For all pairs (a, b) in MM(P, ν4 ) it holds that there is q ∈ B(uab ) such that a 6 q in P , but it is not clear whether there is such an element being also contained in B(pab ). It is the purpose of function α4 to distinguish the two possible cases at this point. Given (a, b) ∈ MM(P, ν4 ), let α4 (a, b) ∈ {yes, no} be the answer to the following question about (a, b): “Is there an element q ∈ B(uab ) ∩ B(pab ) with a 6 q in P ?”. Before defining the function α5 we first show some useful properties of pairs in MM(P, ν5 , Σ) for Σ ∈ Σ(ν5 ). Note that these pairs (a, b) satisfy α2 (a, b) = yes and α4 (a, b) = no. Claim 12. Let Σ ∈ Σ(ν5 ) and suppose that {(ai , bi )}ki=1 is an alternating cycle in MM(P, ν5 , Σ). Let ui denote uai bi , for each i ∈ {1, 2, . . . , k}. Then (i) ui and ui+1 are comparable in T for each i ∈ {1, 2, . . . , k}, and (ii) there is an index j ∈ {1, 2, . . . , k} such that uj 6 ui in T for each i ∈ {1, 2, . . . , k}. Proof. Let Σ ∈ Σ(ν5 ) = {(left, yes, no), (right, yes, no)}. Again we may assume Σ = (left, yes, no) as the other case is symmetrical. Thus α1 (ai , bi ) = left for each i ∈ {1, . . . , k}. We denote uai bi , wai bi , pai bi by ui , wi , pi respectively, for each i ∈ {1, 2, . . . , k}. To prove the first item observe that since α4 (ai , bi ) = no for all pairs (ai , bi ), and since ai 6 bi+1 in P , we have ui < bTi+1 in T . Indeed, otherwise the path from aTi to bTi+1 in T would go through ui and pi , and hence ai 6 bi+1 would hit B(ui ) ∩ B(pi ), contradicting α4 (ai , bi ) = no. Clearly ui+1 < bTi+1 in T . Therefore, {ui , ui+1 } < bTi+1 in T , which makes ui and ui+1 comparable in T . The second item follows immediately from the first item and Observation 8.  Thanks to Claim 12 we know that for every Σ ∈ Σ(ν5 ), each alternating cycle in MM(P, ν5 , Σ) can be written as {(ai , bi )}ki=1 in such a way that ua1 b1 6 uai bi in T for i ∈ {1, . . . , k}. We may further assume that the pair (a1 , b1 ) is chosen in such a way that (i) if α1 (a1 , b1 ) = left then bT1 is to the right of bTi in T for each i ∈ {2, . . . , k} satisfying ua1 b1 = uai bi . 12 (ii) if α1 (a1 , b1 ) = right then bT1 is to the left of bTi in T for each i ∈ {2, . . . , k} satisfying ua1 b1 = uai bi . Note that the pair (a1 , b1 ) is uniquely defined; we call it the root of the alternating cycle. Now for each Σ ∈ Σ(ν5 ) and (a, b) ∈ MM(P, ν5 , Σ) we take a closer look at elements in B(uab ). The bag B(uab ) consists of three distinct elements; let us denote them xab , yab , zab . Given that α2 (a, b) = yes and α4 (a, b) = no, we may assume without loss of generality a 6 xab 66 b in P ; (2) B(uab ) ∩ B(pab ) = {yab , zab }. (3) Recall that the uab wab edge lies on the path from r to bT in T . This implies that the relation a0 6 b hits B(uab ) ∩ B(wab ). Clearly, it cannot hit xab , and thus a0 6 b hits at least one of yab , zab . Let us suppose without loss of generality that this is the case for yab . It follows that a 66 yab 6 b in P ; (4) a0 6 yab in P ; (5) a 66 zab in P ; (6) yab ∈ B(uab ) ∩ B(wab ). (7) With these notations, we define α5 in the following way: For each Σ ∈ Σ(ν5 ) and pair (a, b) ∈ MM(P, ν5 , Σ), we let α5 (a, b) := (φ(xab ), φ(yab ), φ(zab )). (Recall that φ(w) is the color of the element w ∈ X in the 3-coloring φ of the intersection graph defined by the subtrees Tx (x ∈ X), and that xab , yab , zab have distinct colors.) Hence there are 6 possible answers for α5 (a, b). In the following when considering nodes νi of Ψ that are descendants of ν5 , all we will need is that min-max pairs (a, b) ∈ MM(P, νi , Σ) have the same value α5 (a, b) but the value itself will not be important. This is why Ψ does not branch at ν5 . 3.4. Node ν6 and its function α6 . Before defining the next function α6 , let us show some useful properties of strict alternating cycles in MM(P, ν6 , Σ) for Σ ∈ Σ(ν6 ). These properties will be used not only when considering the second leaf ν7 of Ψ but also later on when considering the third leaf ν10 . Now, recall that pairs (a, b) ∈ MM(P, ν6 ) satisfy • α2 (a, b) = yes, and hence there is q ∈ B(uab ) with a 6 q in P , • α4 (a, b) = no, and hence there is no q ∈ B(uab ) ∩ B(pab ) such that a 6 q in P , • the elements of B(uab ) can be labeled with xab , yab , zab such that (2)-(7) hold. We need these properties and the mentioned labeling for the following claim. Claim 13. Let Σ ∈ Σ(ν6 ) and suppose that {(ai , bi )}ki=1 is a strict alternating cycle in MM(P, ν6 , Σ) with root (a1 , b1 ). Let ui , wi denote uai bi , wai bi respectively, for each i ∈ {1, 2, . . . , k}. Then u1 < w1 6 uk < bT1 in T . Proof. We denote pai bi , xai bi , yai bi , zai bi by pi , xi , yi , zi respectively, for each i ∈ {1, 2, . . . , k}. We assume that α1 (a, b) = left for each (a, b) ∈ MM(P, ν6 , Σ). In particular, α1 (ai , bi ) = left for each i ∈ {1, 2, . . . , k}. The path from aTk to bT1 in T cannot go through the edge uk pk , since otherwise by Observation 10 the relation ak 6 b1 would hit B(uk ) ∩ B(pk ) which contradicts the fact that α4 (ak , bk ) = no. This implies uk < bT1 in T . Next we prove that u1 6= uk . Suppose to the contrary that u1 = uk . Then uk lies on the paths from aTk to r and from bT1 to r in T , implying that aTk ∧ bT1 > uk in T . If aTk ∧ bT1 = uk then the path from aTk to bT1 in T goes through uk . Thus, the relation ak 6 b1 hits B(uk ) = {xk , yk , zk } and hence ak 6 xk 6 b1 in P (as ak 66 yk and ak 66 zk in P ). Since B(uk ) = B(u1 ) and α5 (a1 , b1 ) = α5 (ak , bk ) implying φ(xk ) = φ(x1 ), we get xk = x1 . Now a1 6 x1 = xk 6 b1 in P gives a contradiction. If aTk ∧ bT1 > uk in T then it follows that vk 6 bT1 in T . By Observation 9 (ii) we conclude that bT1 is left of bTk in T . Since uk = u1 , this contradicts the fact that (a1 , b1 ) is the root of {(ai , bi )}ki=1 . Therefore, u1 6= uk as claimed, and u1 < uk < bT1 in T . Given the definition of w1 and the fact that uk < bT1 in T , we deduce u1 < w1 6 uk < bT1 in T .  13 Claim 14. Let Σ ∈ Σ(ν6 ) and suppose that {(ai , bi )}ki=1 is a strict alternating cycle in MM(P, ν6 , Σ) with root (a1 , b1 ). Let ui , wi denote uai bi , wai bi respectively, for each i ∈ {1, 2, . . . , k}. Then u1 < w1 6 ui in T for each i ∈ {2, . . . , k}. Proof. We denote vai bi , pai bi , xai bi , yai bi , zai bi by vi , pi , xi , yi , zi respectively, for each i ∈ {1, 2, . . . , k}. We assume that α1 (a, b) = left for each (a, b) ∈ MM(P, ν6 , Σ). In particular, α1 (ai , bi ) = left for each i ∈ {1, 2, . . . , k}. By Claim 13 we have w1 6 uk in T . Arguing by contradiction, suppose that w1 66 ui for some i ∈ {2, . . . , k − 1}, and let i be the largest such index. Thus, w1 6 ui+1 in T . Note also that in this case we must have k > 3. Since ui and ui+1 are comparable in T (by Claim 12) and u1 is minimal in T among all the ui ’s, we obtain u1 = ui < w1 6 ui+1 in T . Observe that ui 6 aTi ∧ bTi+1 in T , as ui < {aTi , bTi+1 } in T . If ui = aTi ∧ bTi+1 then the path from T ai to bTi+1 in T goes through ui . Thus, the relation ai 6 bi+1 hits B(ui ) = {xi , yi , zi }, and it follows that ai 6 xi 6 bi+1 in P (as ai yi and ai zi ). Since B(ui ) = B(u1 ) and φ(xi ) = φ(x1 ) (because α5 (a1 , b1 ) = α5 (ai , bi )), we deduce that xi = x1 . But then a1 6 x1 = xi 6 bi+1 in P , which is a contradiction. (Recall that i > 2.) If u1 = ui < aTi ∧ bTi+1 in T then we must have w1 6 aTi ∧ bTi+1 in T , since aTi ∧ bTi+1 has to be an internal node of the path from ui to bTi+1 in T and since w1 is the neighbor of ui on that path (as ui < w1 6 ui+1 < bTi+1 in T ). In particular, this implies ui < w1 < aTi in T , and it follows that vi = w1 . As aTi is left of bTi in T , and since vi = w1 < bT1 in T , by Observation 9 (ii) we obtain that bT1 is left of  bTi , which contradicts the fact that (a1 , b1 ) is the root of {(ai , bi )}ki=1 . This completes the proof. Now let us define the function α6 . We set α6 (a, b) to be the answer to the following question: “Is B(uab ) ∩ B(wab ) = {xab , yab }?”. Note that yab always belongs to the intersection, and hence α6 tells us whether xab or zab is the other element in B(uab ) ∩ B(wab ). If the answer to this question is “no”, then our signature tree leads us to the second leaf of Ψ, leaf ν7 . 3.5. Second leaf of Ψ: the node ν7 . In this section we show that incomparable pairs in MM(P, ν7 , Σ) are reversible for each Σ ∈ Σ(ν7 ). Recall that MM(P, ν7 ) is a subset of MM(P, ν6 ), allowing us to use the observations and claims from the previous section. Moreover, for every pair (a, b) ∈ MM(P, ν7 ) we have that α6 (a, b) = no, implying that B(uab ) ∩ B(wab ) = {yab , zab }. Claim 15. MM(P, ν7 , Σ) is reversible for each Σ ∈ Σ(ν7 ). Proof. Let Σ ∈ Σ(ν7 ). We assume that α1 (a, b) = left for each (a, b) ∈ MM(P, ν7 , Σ). In particular, α1 (ai , bi ) = left for each i ∈ {1, 2, . . . , k}. Arguing by contradiction, suppose that there is a strict alternating cycle {(ai , bi )}ki=1 in MM(P, ν7 , Σ) with root (a1 , b1 ). We have u1 < w1 6 u2 in T by Claim 14, and in particular aT1 ∧bT2 = u1 . Thus, the path from aT1 to bT2 in T includes the edge u1 w1 . Hence, the relation a1 6 b2 hits B(u1 ) ∩ B(w1 ) ⊆ {x1 , y1 , z1 }. Since a1 6 x1 and a1 k{y1 , z1 } in P we obtain x1 ∈ B(u1 ) ∩ B(w1 ). Recalling that we also have y1 ∈ B(u1 ) ∩ B(w1 ), it follows that B(u1 ) ∩ B(w1 ) = {x1 , y1 }, contradicting α6 (a1 , b1 ) = no.  3.6. Node ν8 and its function α8 . Before we study strict alternating cycles in MM(P, ν8 ), let us recall some useful properties of incomparable pairs in MM(P, ν8 ). Each pair (a, b) ∈ MM(P, ν8 ) satisfies • α4 (a, b) = no, and hence there is no q ∈ B(uab ) ∩ B(pab ) such that a 6 q in P , • the elements of B(uab ) can be labeled with xab , yab , zab such that (2)-(7) hold, • α6 (a, b) = yes, and hence B(uab ) ∩ B(wab ) = {xab , yab }. We proceed with an observation about MM(P, ν8 , Σ) for fixed Σ ∈ Σ(ν8 ). Claim 16. Let Σ ∈ Σ(ν8 ) and suppose that {(ai , bi )}ki=1 is a strict alternating cycle in MM(P, ν8 , Σ) with root (a1 , b1 ). Let ui denote uai bi for each i ∈ {1, 2, . . . , k}. Then u1 < w1 6 u2 < bT1 in T . Proof. We denote wai bi , pai bi , xai bi , yai bi , zai bi by wi , pi , xi , yi , zi respectively, for each i ∈ {1, 2, . . . , k}. We assume that α1 (a, b) = left for each (a, b) ∈ MM(P, ν8 , Σ). In particular, α1 (ai , bi ) = left for each i ∈ {1, 2, . . . , k}. 14 bTi aT2 bT2 ui−1 u2 aT1 bT1 wi ui w1 u1 r Figure 7. A possible situation in the proof of Claim 16. If k = 2 then the claim follows from Claim 13. So we assume k > 3 from now on. By Claim 14 we already know u1 < w1 6 u2 in T , and thus it remains to show u2 < bT1 in T . Arguing by contradiction suppose that u2 6< bT1 in T . Let i ∈ {3, . . . , k} be smallest such that ui u2 in T . There is such an index since uk < bT1 in T by Claim 13, and thus uk u2 in T . See Figure 7 for an illustration of this and upcoming arguments. By our choice of i, we have u1 < u2 6 ui−1 in T . Note that ai−1 6 bi in P , which combined with α4 (ai−1 , bi−1 ) = no yields ui−1 < bTi in T . Hence u2 6 ui−1 < bTi in T . Since u2 < bTi and ui < bTi in T , the two nodes u2 and ui are comparable in T , and thus ui < u2 since ui u2 in T . Combining this with u2 < bTi in T we further deduce that wi 6 u2 in T . On the other hand, w1 6 ui in T by Claim 14. To summarize, we have u1 < w1 6 ui < wi 6 u2 < bTi in T . Now consider the path from aT1 to bT2 in T . Since w1 < u2 6 bT2 and w1 aT1 in T , this path goes through u1 , and thus includes the edge ui wi . Hence the relation a1 6 b2 hits the set B(ui ) ∩ B(wi ), the latter being equal to {xi , yi } since α6 (ai , bi ) = yes. Therefore, a1 6 xi 6 b2 or a1 6 yi 6 b2 in P . But this implies ai 6 xi 6 b2 or a1 6 yi 6 bi in P , a contradiction in both cases to the properties of a strict alternating cycle (recall that i ∈ {3, . . . , k}).  Given Σ ∈ Σ(ν8 ), we say that pairs (a, b), (a′ , b′ ) ∈ MM(P, ν8 , Σ) form a special 2-cycle if, exchanging (a, b) and (a′ , b′ ) if necessary, we have (i) a 6 b′ and a′ 6 b in P , and (ii) uab < wab 6 ua′ b′ < wa′ b′ < bT in T . The first requirement is simply that (a, b), (a′ , b′ ) is a (strict) alternating cycle. (Note that every alternating cycle of length 2 is strict.) As a consequence of this and Claim 16, we know that the first three inequalities of the second requirement are satisfied. So the question here is whether also wa′ b′ < bT holds in T . An implication of the second requirement is that the paths from aT to b′T and from a′T to bT in T both go through the edge ua′ b′ wa′ b′ of T . Note also that the pair (a, b) is the root this special 2-cycle. Let SΣ be the graph with vertex set MM(P, ν8 , Σ) where distinct pairs (a, b), (a′ , b′ ) ∈ MM(P, ν8 , Σ) are adjacent if and only if they form a special 2-cycle. Claim 17. The graph SΣ is bipartite for each Σ ∈ Σ(ν8 ). Proof. Arguing by contradiction, suppose that there is an odd cycle C = {(ai , bi )}ki=1 in SΣ for some Σ ∈ Σ(ν8 ). We may assume that C is induced. Let ui := uai bi , wi := wai bi , xi := xai bi and yi := yai bi for each i ∈ {1, . . . , k}. First we consider the case k = 3. Since u1 , u2 , u3 are pairwise comparable in T , we may assume that u1 < u3 < u2 in T (recall that consecutive ui ’s are distinct by property (ii) of special 2-cycles). By the definition of special 2-cycles, we then obtain u2 < w2 < {bT1 , bT2 , bT3 } in T . Thus the paths from aT1 to bT2 , from aT2 to bT3 , and from aT3 to bT1 in T all go through the edge u2 w2 . This implies that two relations must hit the same element and therefore there is i ∈ {1, 2, 3} and q ∈ B(u2 ) ∩ B(w2 ) such that ai 6 q 6 bi+1 and ai+1 6 q 6 bi+2 in P (indices are taken cyclically). However, this gives ai+1 6 bi+1 in P , a contradiction. 15 Next consider the case k > 5. We will show that C has a chord, contradicting the fact that C is induced. (We remark that the parity of k will not be used here, only that k > 5.) We may suppose that u2 is maximal in T among all the ui ’s. We may also assume without loss of generality u1 6 u3 < u2 in T . (Recall that by property (ii) of special 2-cycles ui and ui+1 are comparable in T and distinct for each i ∈ {1, . . . , k}.) Let i, j be such that {i, j} = {3, 4} and uj < ui in T . (Note that uj 6= ui since (aj , bj ), (ai , bi ) form a special 2-cycle.) We claim that wj < wi 6 w2 and w1 6 wi (8) in T . The inequality wj < wi follows from the fact that (aj , bj ) and (ai , bi ) form a special 2-cycle, and thus in particular uj < wj 6 ui < wi in T . For the inequality wi 6 w2 we do a case distinction. If i = 3 then (a2 , b2 ) and (ai , bi ) form a special 2-cycle with ui < wi 6 u2 < w2 in T . If i = 4 then (a2 , b2 ), (a3 , b3 ) as well as (a3 , b3 ), (ai , bi ) form special 2-cycles. In particular, {w2 , wi } < bT3 in T . This makes w2 and wi comparable in T and by the choice of u2 we have wi 6 w2 , as desired. Besides this, we also have w1 < w2 in T (as (a1 , b1 ), (a2 , b2 ) form a special 2-cycle), which makes w1 and wi comparable in T . Since we have u1 6 ui by our choice of i and by the assumption that u1 6 u3 in T , it follows that w1 6 wi in T and (8) is proven. Now, we are going to argue that the aT1 –bT2 path, the a0 –bT1 path, the aTj –bTi path, and the aTi –bTj path all go through the edge ui wi in T . For the aT1 –bT2 path note that u1 < w1 6 wi 6 w2 < bT2 in T . This implies that this path has to go first from aT1 down to u1 and then pursue with the u1 –bT2 path, which includes wi by the previous inequalities, and thus also its parent ui (since u1 < wi in T ). Hence, it includes the edge ui wi of T . For the a0 –bT1 path it suffices to observe that r < ui < wi 6 w2 < bT1 in T . Similarly, for the aTj –bTi path, notice that uj < wj 6 ui < wi < bTi in T . Finally, for the aTi –bTj path, observe that ui < wi < bTj in T . Using Observation 10, it follows that the relations a1 6 b2 , a0 6 b1 , aj 6 bi and ai 6 bj in P all hit B(ui ) ∩ B(wi ) = {xi , yi }. Clearly, a j 6 y i 6 bi and a i 6 x i 6 bj in P . Now, in P we either have a0 6 xi 6 b1 and a1 6 yi 6 b2 , or a0 6 yi 6 b1 and a1 6 xi 6 b2 . This implies ai 6 b1 and a1 6 bi , or aj 6 b1 and a1 6 bj . In the first case, (a1 , b1 ), (ai , bi ) is an alternating cycle of length 2 (and thus is strict). Recall that w2 < bT1 in T , which together with (8) yields w1 6 wi < bT1 in T . Furthermore, applying Claim 14 on (a1 , b1 ), (ai , bi ) we obtain u1 6= ui , implying w1 6= wi , and hence w1 < wi in T . It follows that u1 < w1 6 ui < wi < bT1 in T , which shows that (a1 , b1 ), (ai , bi ) is a special 2-cycle. This gives us a chord of the cycle C, a contradiction. In the second case, (a1 , b1 ), (aj , bj ) is a (strict) alternating cycle. Moreover, w1 6 wi and wj 6 wi in T (see (8)), which makes w1 and wj comparable in T . Again by Claim 14 we get w1 6= wj . If w1 < wj then it also holds that u1 < w1 6 uj < wj < bT1 in T (as wj 6 w2 < bT1 ). If wj < w1 then it follows that uj < wj 6 u1 < w1 < bTj in T (as w1 6 wi < bTj ). Thus in both cases (a1 , b1 ), (aj , bj ) is a special 2-cycle and a chord in C, a contradiction. This completes the proof.  Using Claim 17, for each Σ ∈ Σ(ν8 ) let ψ8,Σ : MM(P, ν8 , Σ) → {1, 2} be a (proper) 2-coloring of SΣ . The function α8 then simply records the color of a pair in this coloring: For each Σ ∈ Σ(ν8 ) and each pair (a, b) ∈ MM(P, ν8 , Σ), we let α8 (a, b) := ψ8,Σ (a, b). 3.7. Node ν9 and its function α9 . By the definition of α8 there is no special 2-cycle in MM(P, ν9 , Σ), for every Σ ∈ Σ(ν9 ). This will be used in the definition of the function α9 . In order to define α9 we first need to introduce an auxiliary directed graph. For each Σ ∈ Σ(ν9 ), let KΣ be the directed graph with vertex set MM(P, ν9 , Σ) where for any two distinct pairs (a1 , b1 ), (a2 , b2 ) ∈ MM(P, ν9 , Σ) there is an arc from (a1 , b1 ) to (a2 , b2 ) in KΣ if there exists a strict alternating cycle {(a′i , b′i )}ki=1 in MM(P, ν9 , Σ) with root (a′1 , b′1 ) such that (a′1 , b′1 ) = (a1 , b1 ) and (a′2 , b′2 ) = (a2 , b2 ). In that case we say that the arc f is induced by the strict alternating cycle {(a′i , b′i )}ki=1 . (Note that there could possibly be different strict alternating cycles inducing the same arc f .) 16 Claim 18. Let Σ ∈ Σ(ν9 ). Then for each arc ((a1 , b1 ), (a2 , b2 )) in KΣ we have (i) x1 6 y2 , and (ii) y1 6 z2 6 b1 in P , where xi := xai bi , yi := yai bi , and zi := zai bi for i = 1, 2. Proof. Let ui := uai bi , pi := pai bi , and wi := wai bi for i = 1, 2. By the definition of an arc in KΣ and by Claim 16 it holds that u1 < w1 6 u2 < bT1 in T . Thus, the path from aT1 to bT2 in T goes through u1 , w1 , u2 and w2 . Hence, the relation a1 6 b2 (which exists by the definition of an arc in KΣ ) hits B(u1 ) ∩ B(w1 ) = {x1 , y1 } and B(u2 ) ∩ B(w2 ) = {x2 , y2 }. It cannot hit y1 (otherwise a1 6 y1 6 b1 in P ) nor x2 (otherwise a2 6 x2 6 b2 in P ). It follows that a1 6 x1 6 y2 6 b2 in P by Observation 10, and (i) is proven. For (ii) observe that the relation a0 6 b1 hits {x1 , y1 } = B(u1 ) ∩ B(w1 ) and {y2 , z2 } = B(p2 ) ∩ B(u2 ). It cannot hit x1 nor y2 since otherwise a1 6 b1 in P as we have a1 6 x1 6 y2 in P . Hence we obtain a0 6 y1 6 z2 6 b1 in P by Observation 10.  Let us remark that it is possible to strengthen the statement of Claim 18. In fact, strict inequalities x1 < y2 and y1 < z2 hold in (i) and (ii), respectively. Indeed, since φ is a proper coloring and α5 (a1 , b1 ) = α5 (a2 , b2 ), we have φ(x1 ) = φ(x2 ) 6= φ(y2 ) and φ(y1 ) = φ(y2 ) 6= φ(z2 ), implying that x1 6= y2 and y1 6= z2 . For our purposes the non-strict inequalities of Claim 18 (and also in forthcoming claims) suffice. As we do not want give more arguments than needed, we will not always prove the strongest possible statement in the following. Claim 19. Let Σ ∈ Σ(ν9 ). Suppose that (a1 , b1 ), (a2 , b2 ), (a3 , b3 ) ∈ MM(P, ν9 , Σ) are three distinct pairs such that ((a1 , b1 ), (a2 , b2 )) is an arc in KΣ and u1 < u3 < w3 6 u2 in T , where ui := uai bi and wi := wai bi for each i ∈ {1, 2, 3}. Then (i) x1 6 z3 6 y2 , and (ii) y1 6 y3 6 z2 6 b1 in P , where xi := xai bi , yi := yai bi , and zi := zai bi for each i ∈ {1, 2, 3}. Proof. Let pi := pai bi for each i ∈ {1, 2, 3}. We have x1 6 y2 and y1 6 z2 6 b1 in P by Claim 18. Since u1 < u3 < w3 6 u2 in T , the relations x1 6 y2 and y1 6 z2 hit B(u3 ) ∩ B(w3 ) = {x3 , y3 }. They cannot hit the same element since otherwise a1 6 x1 6 z2 6 b1 in P . If x1 6 y3 6 y2 in P then y1 6 x3 6 z2 , and we conclude a1 6 x1 6 y3 6 b3 and a3 6 x3 6 z2 6 b1 in P . Thus, (a1 , b1 ) and (a3 , b3 ) form an alternating cycle of length 2. Applying Claim 16 on the pairs (a1 , b1 ), (a2 , b2 ) we obtain in particular u2 < bT1 in T . Together with our assumptions it follows that both u3 and w3 lie on the path from u1 to bT1 in T . Hence, u1 < w1 6 u3 < w3 < bT1 in T , and therefore (a1 , b1 ), (a3 , b3 ) is a special 2-cycle, contradicting the fact that ψ8,Σ (a1 , b1 ) = α8 (a1 , b1 ) = α8 (a3 , b3 ) = ψ8,Σ (a3 , b3 ). We conclude that x1 6 x3 6 y2 and y1 6 y3 6 z2 in P . It remains to show that x1 6 z3 6 y2 in P . For this, note that x1 6 x3 hits B(p3 )∩B(u3 ) = {y3 , z3 }. It cannot hit y3 as otherwise a1 6 x1 6 y3 6 z2 6 b1 in P . This implies x1 6 z3 6 x3 6 y2 in P , as desired.  We are now ready to prove our main claim about KΣ (Σ ∈ Σ(ν9 )), namely that KΣ is bipartite. (We consider a directed graph to be bipartite if its underlying undirected graph is.) Claim 20. The graph KΣ is bipartite for each Σ ∈ Σ(ν9 ). Proof. Arguing by contradiction, suppose that there is an odd undirected cycle C = {(ai , bi )}ki=1 in KΣ for some Σ ∈ Σ(ν9 ). Let ui := uai bi , wi := wai bi , xi := xai bi , yi := yai bi , and zi := zai bi for each i ∈ {1, . . . , k}. We may assume that α1 (ai , bi ) = left for each i ∈ {1, . . . , k}. Consider the cyclic sequence of nodes (u1 , u2 , . . . , uk ). It might be the case that some of the nodes coincide. In order to avoid this, we modify the sequence as follows: If ui = uj for some i, j ∈ {1, . . . , k} with i < j, then consider the two cyclic sequences (ui , ui+1 , . . . , uj−1 ) and (uj , uj+1 , . . . , ui−1 ) (thus the second one contains uk , and also u1 if i > 1). Since k is odd, exactly one of the two cyclic sequences has odd length. We replace the original sequence by that one, and repeat this process as long as some node appears at least twice in the current cyclic sequence. We claim that at every stage of the above modification process the cyclic sequence S = (ui1 , ui2 , . . . , uiℓ ) under consideration satisfies the following property: For every s ∈ {1, . . . , ℓ} there is an index j ∈ {1, . . . , k} such that uis = uj and uis+1 = uj+1 (taking indices cyclically in each case, as expected). This property obviously holds at the start, so let us show that it remains true during the 17 bTt+1 bTs uin−1 = ut ui2 = us+1 bTr w = ws ui1 = ur+1 = us u iℓ = u r uin = ut+1 r Figure 8. Possible situation in proof of Claim 20. Note that we could also have ur < ut+1 in T . rest of the procedure. Thus suppose that the current cyclic sequence S = (ui1 , ui2 , . . . , uiℓ ) satisfies the property, and that we modify it because of two indices p, q ∈ {1, . . . , ℓ} with p < q such that uip = uiq . Without loss of generality we may assume that the resulting odd sequence is S ′ = (uip , . . . , uiq−1 ). We only need to show that there is an index j ∈ {1, . . . , k} such that uiq−1 = uj and uip = uj+1 , since uiq−1 and uip are the only two consecutive nodes in S ′ that were not consecutive in S. Then it suffices to take j ∈ {1, . . . , k} such that uiq−1 = uj and uiq = uj+1 , which exists since S satisfies our property, and observe that uip = uiq = uj+1 . Therefore, the property holds at every step, as claimed. Recalling that any two consecutive nodes in the original cyclic sequence (u1 , u2 , . . . , uk ) are distinct and comparable in T (by Claim 16), it follows from the property considered above that this holds at every step of the modification procedure, and thus in particular for the final sequence S = (ui1 , ui2 , . . . , uiℓ ) resulting from the procedure. In particular, ℓ > 3, because the cycle has odd length. Since ℓ is odd, there exists m ∈ {1, . . . , ℓ} such that uim−1 < uim < uim+1 or uim−1 > uim > uim+1 in T . Reversing the ordering of C and the cyclic sequence S if necessary, we may assume without loss of generality uim−1 < uim < uim+1 in T . Similarly, shifting the sequence S cyclically if necessary, we may assume m = 1. Thus uiℓ < ui1 < ui2 in T . Let w be the neighbor of ui1 on the ui1 –ui2 path in T . Thus w 6 ui2 in T . Now let n ∈ {3, . . . , ℓ} be minimal such that w uin in T . Since uiℓ < w in T , this index exists. As uin−1 and uin are comparable in T , it follows that uin < w 6 uin−1 in T . (This follows from the definition of n if n > 3, and from the fact that w 6 ui2 in T if n = 3.) Furthermore we have uin 6= ui1 , because n 6= 1 and all nodes in S are distinct. We conclude that (9) uin < ui1 < w 6 uin−1 in T . Now, by the property that is maintained during the modification process of S, there exist indices r, s, t ∈ {1, . . . , k} such that u iℓ = u r and ui1 = ur+1 , (10) u i1 = u s and ui2 = us+1 , (11) uin−1 = ut and uin = ut+1 . (12) It follows that KΣ contains the following arcs: ((ar , br ), (ar+1 , br+1 )), ((as , bs ), (as+1 , bs+1 )), and ((at+1 , bt+1 ), (at , bt )). Applying Claim 16 on these arcs we obtain: ur < wr 6 ur+1 < bTr , (13) us < ws 6 us+1 < bTs , (14) bTt+1 (15) ut+1 < wt+1 6 ut < in T . See Figure 8 for a possible configuration in T and for upcoming arguments. Since ui1 = us and ui2 = us+1 we have w = ws . Hence with (9) we get ut+1 < us < ws 6 ut in T , and by Claim 19 it follows that xt+1 6 zs 6 yt and yt+1 6 ys 6 zt 6 bt+1 in P . Now applying Claim 18 on the arc ((ar , br ), (ar+1 , br+1 )) we get xr 6 yr+1 and 18 yr 6 zr+1 6 br in P . Since α5 (ar+1 , br+1 ) = α5 (as , bs ) and B(ur+1 ) = B(us ) (because ur+1 = us ), we conclude that yr+1 = ys and zr+1 = zs . Using this and the derived inequalities we see that at+1 6 xt+1 6 zs = zr+1 6 br and ar 6 xr 6 yr+1 = ys 6 bt+1 in P . Thus, (ar , br ) and (at+1 , bt+1 ) form an alternating cycle of length 2. In particular, this shows r 6= t + 1 (otherwise we would have ar 6 bt+1 = br in P ), and consequently (by Claim 16) ur 6= ut+1 . (16) Observe that ur < ur+1 = ui1 (by (13) and (10)) and ut+1 = uin < ui1 (by (12) and (9)) in T , which makes ur and ut+1 comparable in T . Furthermore, (10) (13) ui1 = ur+1 < bTr and (9) (12) (15) ui1 < uin−1 = ut < bTt+1 in T , so all together we have {ur , ut+1 } < ui1 < {bTr , bTt+1 } in T . But from this and (16) it follows that either ur < wr 6 ut+1 < wt+1 6 ui1 < bTr or ut+1 < wt+1 6 ur < wr 6 ui1 < bTt+1 holds in T . Both cases imply that (ar , br ) and (at+1 , bt+1 ) form a special 2-cycle, which is our final contradiction.  Using Claim 20 we let ψ9,Σ : MM(P, ν9 , Σ) → {1, 2} be a 2-coloring of KΣ , for each Σ ∈ Σ(ν9 ). The function α9 then records the color of a pair in this coloring: For each Σ ∈ Σ(ν9 ) and each pair (a, b) ∈ MM(P, ν9 , Σ), we let α9 (a, b) := ψ9,Σ (a, b). 3.8. Third leaf of Ψ: ν10 . Now suppose that there was a strict alternating cycle {(ai , bi )}ki=1 with root (a1 , b1 ) in MM(P, ν10 , Σ), for some Σ ∈ Σ(ν10 ). Then, by the definition of KΣ , there is an arc from (a1 , b1 ) to (a2 , b2 ) in KΣ , and hence α9 (a1 , b1 ) 6= α9 (a2 , b2 ), a contradiction. Therefore, there is no such cycle in MM(P, ν10 , Σ), and we have established the following claim: Claim 21. The set MM(P, ν10 , Σ) is reversible for each Σ ∈ Σ(ν10 ). This concludes our study of the third leaf of Ψ. 3.9. Node ν11 and its functions α11 . In this section we consider pairs (a, b) ∈ MM(P, ν11 ). Recall that in this case (a, b) satisfies α2 (a, b) = yes and even stronger α4 (a, b) = yes, and hence there is q ∈ B(uab ) ∩ B(pab ) such that a 6 q in P . For every such pair (a, b), denote the three elements in B(uab ) as xab , yab , zab in such a way that B(uab ) ∩ B(pab ) = {xab , yab } (17) a 6 xab 66 b; (18) a 66 yab 6 b (19) and in P . (Here we use that α4 (a, b) = yes and that a0 6 b hits B(uab ) ∩ B(pab ).) The function α11 is defined similarly as α5 in Section 3.5: For each Σ ∈ Σ(ν11 ) and pair (a, b) ∈ MM(P, ν11 , Σ), we let α11 (a, b) := (φ(xab ), φ(yab ), φ(zab )). (Recall that φ is the 3-coloring of X defined earlier on which is such that x, y ∈ X receive distinct colors whenever V (Tx ) ∩ V (Ty ) 6= ∅; in particular, the three colors φ(xab ), φ(yab ), φ(zab ) are distinct.) 19 3.10. Node ν12 and its functions α12 . We start by defining function α12 : For each Σ ∈ Σ(ν12 ) and pair (a, b) ∈ MM(P, ν12 , Σ), we let α12 (a, b) record the three yes/no answers to three independent questions about the pair (a, b), namely “Is a 6 zab in P ?”; (Q1) “Is zab 6 b in P ?”; (Q2) “Is a0 6 xab in P ?”. (Q3) (Recall that a0 < b in P for all b ∈ max(P )). Formally speaking, α12 (a, b) is defined as the vector (s1 , s2 , s3 ) ∈ {yes, no}3 where si is the answer to the i-th question above, for i = 1, 2, 3. Note that we cannot have a 6 zab and zab 6 b at the same time in P , and thus there are only 6 possible vectors of answers. (This is why the corresponding edge in the signature tree Ψ is labeled 6 instead of 8.) 3.11. Node ν13 and its functions α13 : Dealing with strict alternating cycles of length 2. Let us first summarize the properties of pairs in MM(P, ν13 ). For each Σ ∈ Σ(ν13 ) and each (a, b) ∈ MM(P, ν13 , Σ) we have that • elements of B(uab ) can be labeled with xab , yab , zab such that (17)-(19) hold, • φ(xab ), φ(yab ), φ(zab ) are pairwise distinct, • φ(xab ) = φ(xa′ b′ ), φ(yab ) = φ(ya′ b′ ) and φ(zab ) = φ(za′ b′ ) for every (a′ , b′ ) ∈ MM(P, ν13 , Σ), • all pairs of MM(P, ν13 , Σ) produce the same answers to (Q1)-(Q3). For each Σ ∈ Σ(ν13 ) let JΣ be the graph with vertex set MM(P, ν13 , Σ) where two distinct pairs (a, b), (a′ , b′ ) ∈ MM(P, ν13 , Σ) are adjacent if and only if (a, b), (a′ , b′ ) is an alternating cycle. Our goal in this section is to show that JΣ is 4-colorable: Lemma 22. For each Σ ∈ Σ(ν13 ) there is a proper coloring of JΣ with 4 colors. To this aim, we show a number of properties of 2-cycles in MM(P, ν13 , Σ). Claim 23. Let Σ ∈ Σ(ν13 ) and suppose that (a1 , b1 ), (a2 , b2 ) is a 2-cycle in MM(P, ν13 , Σ). Let ui := uai bi for i = 1, 2. Then u1 6= u2 . Proof. Let xi := xai bi , yi := yai bi , and zi := zai bi for i = 1, 2. We may assume α1 (ai , bi ) = left for i = 1, 2. Arguing by contradiction suppose that u1 = u2 . Exchanging (a1 , b1 ) and (a2 , b2 ) if necessary we may assume that bT1 is left of bT2 in T . Since in T the node aT1 is left of bT1 , which itself is left of bT2 , the path connecting aT1 to bT2 in T goes through u1 . Thus, the relation a1 6 b2 hits B(u1 ) = {x1 , y1 , z1 } = B(u2 ) = {x2 , y2 , z2 }, and hence a1 6 c 6 b2 in P for some element c ∈ B(u1 ). Recall that ai 6 xi and ai yi in P for i = 1, 2. Thus c ∈ {x1 , z1 }. Moreover, (φ(x1 ), φ(y1 ), φ(z1 )) = (φ(x2 ), φ(y2 ), φ(z2 )) since α11 (a1 , b1 ) = α11 (a2 , b2 ), which implies x1 = x2 , y1 = y2 , and z1 = z2 . If c = x1 then a2 6 x2 = x1 6 b2 in P , a contradiction. If c = z1 then, using that a2 6 z2 in P (since α12 (a1 , b1 ) = α12 (a2 , b2 )), we obtain a2 6 z2 = z1 6 b2 in P , again a contradiction.  By Claim 23 if (a1 , b1 ), (a2 , b2 ) is a 2-cycle in MM(P, ν13 , Σ) for some Σ ∈ Σ(ν13 ) then ua1 b1 6= ua2 b2 . Let us say that the 2-cycle is of type 1 if the latter two nodes are comparable in T (that is, ua1 b1 < ua2 b2 or ua1 b1 > ua2 b2 in T ), and of type 2 otherwise. By extension, each edge of the graph JΣ is either of type 1 or of type 2. Let JΣ,i denote the spanning subgraph of JΣ defined by the edges of type i, for i = 1, 2. Thus JΣ,1 and JΣ,2 are edge disjoint, and JΣ = JΣ,1 ∪ JΣ,2 . In what follows we will first show that JΣ,1 is bipartite, and then considering a 2-coloring of JΣ,1 , we will prove that the two subgraphs of JΣ,2 induced by the two color classes are bipartite. This clearly implies our main lemma, Lemma 22, that JΣ is 4-colorable. Claim 24. Let Σ ∈ Σ(ν13 ) and suppose that (a1 , b1 ), (a2 , b2 ) is a 2-cycle in MM(P, ν13 , Σ) of type 1. Let ui := uai bi and wi := wai bi for i = 1, 2, and suppose further that u1 < u2 in T . Then u1 < w1 6 u2 in T. Proof. Arguing by contradiction, suppose that w1 u2 in T . Let vi := vai bi , pi := p(ui ), xi := xai bi , yi := yai bi , and zi := zai bi for i = 1, 2. First suppose that v1 6 u2 in T . Then the path from aT2 to bT1 in T goes through u2 , p2 , v1 and u1 . Thus the relation a2 6 b1 hits B(u2 ) ∩ B(p2 ) = {x2 , y2 } and B(v1 ) ∩ B(u1 ). Note that the two edges 20 u2 p2 and v1 u1 may coincide (if u2 = v1 ). Since a2 6 b1 cannot hit y2 because y2 6 b2 in P , it hits x2 , and by Observation 10 we then have a 2 6 x 2 6 c 6 b1 (20) in P for some c ∈ B(v1 ) ∩ B(u1 ). Let d be the element in (B(v1 ) ∩ B(u1 )) \ {c}. The aT1 –u1 path and the r–bT2 path in T both go through the edge v1 u1 . Thus, the relations a1 6 x1 and a0 6 b2 both hit {c, d}. Neither can hit c since otherwise a1 6 c 6 b1 or a2 6 c 6 b2 in P . Hence, a1 6 d 6 x1 and a0 6 d 6 b2 in P , which implies a0 6 x1 . We then have a0 6 x2 in P as well, since by α12 (a1 , b1 ) = α12 (a2 , b2 ) both pairs (a1 , b1 ), (a2 , b2 ) give the same answer to question (Q3). Clearly, a0 6 x2 hits {c, d}. This relation cannot hit d since otherwise a1 6 d 6 x2 and x2 6 b1 (by (20)) would imply a1 6 b1 in P . Thus a0 6 c 6 x2 in P . Given that x2 6 c in P by (20), we conclude x2 = c. Using that x1 = c ∈ B(u1 ) = {x1 , y1 , z1 } and (φ(x1 ), φ(y1 ), φ(z1 )) = (φ(x2 ), φ(y2 ), φ(z2 )) (since α11 (a1 , b1 ) = α11 (a2 , b2 )), we further deduce that x1 6= y2 (because φ(x1 ) = φ(x2 ) 6= φ(y2 )) and x1 6= z2 (because φ(x1 ) = φ(x2 ) 6= φ(z2 )) and therefore x2 = c = x1 . However, this implies a1 6 x1 = x2 6 b1 in P , a contradiction. Next assume that v1 u2 in T . Let v1′ be the neighbor of u1 on the u1 –u2 path in T . Thus v1′ 6= w1 and v1′ 6= v1 . The path from aT2 to bT1 in T goes through u2 , p2 , v1′ and u1 . Thus, the relation a2 6 b1 hits B(p2 ) ∩ B(u2 ) = {x2 , y2 } and B(v1′ ) ∩ B(u1 ). It cannot hit y2 since otherwise a2 6 y2 6 b2 in P . By Observation 10, we then have a 2 6 x 2 6 c ′ 6 b1 (21) for some c′ ∈ B(v1′ ) ∩ B(u1 ). Let d′ be the element in (B(v1′ ) ∩ B(u1 )) \ {c′ }. The paths from r to bT2 and from aT1 to bT2 in T both go through u1 and v1′ . Thus the two relations a0 6 b2 and a1 6 b2 hit the set {c′ , d′ }. They cannot hit c′ since otherwise we would get c′ 6 b2 , implying a2 6 c′ 6 b2 in P by (21). Hence, a0 6 d′ 6 b2 and a1 6 d′ 6 b2 in P . Observe that {c′ , d′ } ⊆ {x1 , y1 , z1 } = B(u1 ), and that we have c′ 6= x1 (otherwise a1 6 x1 = c′ 6 b1 in P with (21)) and d′ 6= y1 (otherwise a1 6 d′ = y1 6 b1 in P ). We claim that a0 6 x1 in P . If d′ = x1 this is obvious, so suppose that d′ = z1 , which implies c′ = y1 . The relation a0 6 d′ clearly hits B(p1 ) ∩ B(u1 ) = {x1 , y1 }. If it hits x1 , then a0 6 x1 in P . If, on the other hand, it hits y1 , then together with (21) we obtain a2 6 c′ = y1 6 d′ 6 b2 in P , a contradiction. Hence a0 6 x1 in P , as claimed. We then have a0 6 x2 in P as well, since by α12 (a1 , b1 ) = α12 (a2 , b2 ) both pairs (a1 , b1 ), (a2 , b2 ) give the same answer to question (Q3). The relation a0 6 x2 also hits {c′ , d′ }. It cannot hit d′ , since otherwise a1 6 d′ 6 x2 6 b1 in P by (21). Thus we have a0 6 c′ 6 x2 in P . Note that this yields c′ = x2 , since we had x2 6 c′ in P . Using that c′ ∈ B(u1 ) and (φ(x1 ), φ(y1 ), φ(z1 )) = (φ(x2 ), φ(y2 ), φ(z2 )) (since α11 (a1 , b1 ) = α11 (a2 , b2 )), we deduce that x2 = c′ = x1 . However, this implies a1 6 x1 = x2 6 b1 in P , a contradiction.  Claim 25. Let Σ ∈ Σ(ν13 ) and suppose that (a1 , b1 ), (a2 , b2 ) is a 2-cycle in MM(P, ν13 , Σ) of type 1. Let ui := uai bi , xi := xai bi , and yi := yai bi for i = 1, 2, and suppose further that u1 < u2 in T . Then (i) a1 6 y2 , and (ii) x2 6 b1 in P . Proof. Let pi := p(ui ), vi := vai bi and wi := wai bi for i = 1, 2. The path from aT1 to bT2 in T has to go through u1 since u1 = p(w1 ), w1 6< aT1 and w1 6 u2 < bT2 in T (by Claim 24). As a consequence, this path goes through p2 and u2 . Hence the relation a1 6 b2 hits B(p2 ) ∩ B(u2 ) = {x2 , y2 }. It cannot hit x2 , since otherwise a2 6 x2 6 b2 in P . Therefore, it hits y2 , showing (i). To show (ii), observe that in T at least one of the r–bT1 path and the aT2 –bT1 path goes through p2 and u2 . Thus, at least one of the two relations a0 6 b1 and a2 6 b1 hits {x2 , y2 }. Neither can hit y2 since otherwise a1 6 y2 6 b1 in P . Therefore, x2 6 b1 in P .  Claim 26. The graph JΣ,1 is triangle-free for each Σ ∈ Σ(ν13 ). Proof. Let Σ ∈ Σ(ν13 ). Arguing by contradiction, suppose that there is a triangle (a1 , b1 ), (a2 , b2 ), (a3 , b3 ) in JΣ,1 . Let ui := uai bi , pi := p(ui ), wi := wai bi , xi := xai bi , and yi := yai bi for each i ∈ {1, 2, 3}. Since the nodes u1 , u2 , u3 are pairwise comparable in T and are all distinct (by Claim 23), we may assume without loss of generality u1 < u2 < u3 in T . First we show that a0 6 xi holds in P for some i ∈ {1, 2, 3}. Suppose this is not the case. Consider the path from r to bT3 in T . This path goes through the nodes p1 , u1 , p2 , u2 , p3 and u3 . Hence, the relation 21 a0 6 b3 hits {x1 , y1 }, {x2 , y2 } and {x3 , y3 }. By our assumption it hits yi for each i ∈ {1, 2, 3}, and we have a0 6 y1 6 y2 6 y3 in P by Observation 10. If u2 6 bT1 in T then a0 6 b1 hits {x2 , y2 }, and thus hits y2 by our assumption. Hence y2 6 b1 in P , which using Claim 25 (i) implies a1 6 y2 6 b1 in P , a contradiction. Therefore, u2 k bT1 in T . The fact that u1 < u2 < u3 in T further implies u1 < w 1 6 u2 < w 2 6 u3 by Claim 24. Observe that the aT1 –bT2 path, the aT2 –bT3 path, and the aT3 –bT1 path in T all include the edge u2 w2 . This is clear for the first two paths, and follows from u2 k bT1 in T for the third one. Thus the three relations a1 6 b2 , a2 6 b3 , a3 6 b1 all hit B(u2 ) ∩ B(w2 ) =: {c, d}. Without loss of generality we have a1 6 c 6 b2 in P . This implies a2 6 d 6 b3 in P , which in turn implies a3 6 c 6 b1 . However, it follows that a1 6 c 6 b1 in P , a contradiction. This shows that a0 6 xi holds in P for some i ∈ {1, 2, 3}, as claimed. Now, since α12 (a1 , b1 ) = α12 (a2 , b2 ) = α12 (a3 , b3 ), it follows that a0 6 xi in P for each i ∈ {1, 2, 3}. Consider the relation a0 6 x3 in P . The path from r to u3 in T goes through p2 and u2 . Thus, a0 6 x3 hits {x2 , y2 }. It cannot hit x2 because otherwise a2 6 x2 6 x3 in P , which together with x3 6 b2 (by Claim 25 (ii)) implies a2 6 x3 6 b2 in P . Hence a0 6 x3 hits y2 , and we have y2 6 x3 in P . On the other hand, by Claim 25 we have a1 6 y2 and x3 6 b1 in P . It follows that a1 6 y2 6 x3 6 b1 in P , a contradiction. This concludes the proof.  Claim 27. Let Σ ∈ Σ(ν13 ). Suppose that (a1 , b1 ), (a2 , b2 ), (a3 , b3 ) ∈ MM(P, ν13 , Σ) are three distinct pairs such that (a1 , b1 ), (a2 , b2 ) form a 2-cycle of type 1 and (a1 , b1 ), (a3 , b3 ) do not form a 2-cycle (so neither of type 1 nor of type 2). Let ui := uai bi , xi := xai bi , and yi := yai bi for each i ∈ {1, 2, 3}. Assume further that u1 < u3 6 u2 in T . Then (i) a1 6 x3 , and (ii) y1 6 y3 6 b1 in P . Proof. Let pi := p(ui ) and wi := wai bi for each i ∈ {1, 2, 3}. Since u1 < u3 6 u2 in T and also u1 < w1 6 u2 in T by Claim 24, it follows that u1 < w 1 6 u3 6 u2 in T . First suppose that u3 6 bT1 in T . Then the relation y1 6 b1 hits B(p3 ) ∩ B(u3 ) = {x3 , y3 }. By Claim 25 (i) we have a1 6 y2 in P . Furthermore, a1 6 y2 also hits {x3 , y3 }. Clearly, y1 6 b1 and a1 6 y2 cannot hit the same element of {x3 , y3 }. If y1 6 x3 6 b1 then a1 6 y3 6 y2 in P , which implies a3 6 x3 6 b1 and a1 6 y3 6 b3 , that is, that (a1 , b1 ), (a3 , b3 ) is a 2-cycle, a contradiction. Hence we have y1 6 y3 6 b1 and a1 6 x3 6 y2 in P , as claimed. Next assume that u3 bT1 in T . Then the path from aT2 to bT1 in T includes the edge u3 p3 , and hence a2 6 b1 hits B(p3 ) ∩ B(u3 ) = {x3 , y3 }. The relation a1 6 b2 also hits {x3 , y3 }. Clearly, a2 6 b1 and a1 6 b2 cannot hit the same element of {x3 , y3 }. If a2 6 x3 6 b1 in P then a1 6 y3 6 b2 . However, it then follows a3 6 x3 6 b1 and a1 6 y3 6 b3 in P , implying that (a1 , b1 ), (a3 , b3 ) is a 2-cycle, a contradiction. Therefore, a2 6 y3 6 b1 and a1 6 x3 6 b2 in P . In order to conclude the proof, it only remains to show that y1 6 y3 in P . For this, observe that the path from r to u3 in T includes the edge p1 u1 . Hence, the relation a0 6 y3 hits {x1 , y1 }. It cannot hit x1 since otherwise a1 6 x1 6 y3 6 b1 . Thus, a0 6 y1 6 y3 in P , as desired.  An illustration for the next two claims is given on Figure 9. Claim 28. Let Σ ∈ Σ(ν13 ). Suppose that (a1 , b1 ), (a2 , b2 ), (a3 , b3 ), (a4 , b4 ) ∈ MM(P, ν13 , Σ) are four distinct pairs such that u1 < u4 < u2 and u3 < u4 in T , where ui := uai bi for each i ∈ {1, 2, 3, 4}. Assume further that (a1 , b1 ), (a2 , b2 ) and (a3 , b3 ), (a4 , b4 ) are 2-cycles (which are thus of type 1). Then at least one of (a1 , b1 ), (a3 , b3 ) and (a1 , b1 ), (a4 , b4 ) is a 2-cycle of type 1. Proof. Let xi := xai bi and yi := yai bi for each i ∈ {1, 2, 3, 4}. Suppose that (a1 , b1 ), (a4 , b4 ) is not a 2-cycle, since otherwise we are done. Applying Claim 27 on the three pairs (a1 , b1 ), (a2 , b2 ), (a4 , b4 ) we obtain that a1 6 x4 and y1 6 y4 6 b1 in P . 22 u3 u4 u2 u2 u2 u4 u4 u4 u5 =⇒ u1 u3 or u1 u1 u2 u3 u3 u1 Figure 9. Possible situations in Claim 28 and 29. An edge indicates that its endpoints correspond to the meeting points of a 2-cycle. u3 3 C u2 e′ i := (ai , bi ) u4 2 e E = {{12, 23}, {54, 43}, {71, 12}} T 4 6 1 u6 u1 u5 u7 7 5 r Figure 10. Example for Claim 30: Cycle C on the pairs (a1 , b1 ), . . . , (a7 , b7 ) and the positions of u1 , . . . , u7 in T . Given this, it follows that u{12,23} = u2 is maximal in T among u{12,23} , u{54,43} , u{71,12} . Following the proof for Claim 30 we would get that one of the dotted edges in the figure must be a real edge in JΣ,1 , implying that C cannot be induced. Using Claim 25 on the 2-cycle (a3 , b3 ), (a4 , b4 ) we also obtain a3 6 y4 and x4 6 b3 in P . It follows that a3 6 y4 6 b1 and a1 6 x4 6 b3 in P . Hence (a1 , b1 ), (a3 , b3 ) is a 2-cycle. Furthermore, it is of type 1 because u1 < u4 and u3 < u4 in T , implying that u1 and u3 are comparable in T .  Claim 29. Let Σ ∈ Σ(ν13 ). Suppose that (a1 , b1 ), (a2 , b2 ), (a3 , b3 ), (a4 , b4 ), (a5 , b5 ) ∈ MM(P, ν13 , Σ) are five distinct pairs such that u1 < u5 < u4 and u2 < u5 < u3 in T , where ui := uai bi for each i ∈ {1, . . . , 5}. Assume further that (a1 , b1 ), (a2 , b2 ) and (a2 , b2 ), (a3 , b3 ) and (a1 , b1 ), (a4 , b4 ) are 2-cycles of type 1. Then at least one of (a1 , b1 ), (a5 , b5 ) and (a2 , b2 ), (a5 , b5 ) is a 2-cycle of type 1. Proof. Assume to the contrary that neither (a1 , b1 ), (a5 , b5 ) nor (a2 , b2 ), (a5 , b5 ) is a 2-cycle. (Note that if one is a 2-cycle, then it is automatically of type 1 since u1 < u5 and u2 < u5 in T .) We either have u1 < u2 or u2 < u1 in T . Exploiting symmetry we may assume u1 < u2 in T . (Indeed, if not then it suffices to exchange (a1 , b1 ) and (a4 , b4 ) with respectively (a2 , b2 ) and (a3 , b3 ).) Applying Claim 27 on the three pairs (a1 , b1 ), (a4 , b4 ), (a5 , b5 ) and on the three pairs (a2 , b2 ), (a3 , b3 ), (a5 , b5 ), we obtain y5 6 b1 and y2 6 y5 in P . Since (a1 , b1 ), (a2 , b2 ) is a 2-cycle and u1 < u2 in T , we have a1 6 y2 in P by Claim 25. But all together this implies a1 6 y2 6 y5 6 b1 in P , a contradiction.  Claim 30. The graph JΣ,1 is bipartite for every Σ ∈ Σ(ν13 ). Proof. Let Σ ∈ Σ(ν13 ). Arguing by contradiction, suppose that JΣ,1 is not bipartite. Let C be a shortest odd cycle in JΣ,1 . Thus C is induced, that is, C has no chord. By Claim 26 we know that C has length at least 5. 23 We orient the edges of JΣ,1 in the following natural way: For each edge {(a, b), (a′ , b′ )} in JΣ,1 , we orient the edge towards (a′ , b′ ) if uab < ua′ b′ in T , and towards (a, b) otherwise (that is, if uab > ua′ b′ in T ). We encourage the reader to take a look at Figure 10 for upcoming new notations and arguments. Let E be the set of all pairs of consecutive edges in C such that the source of one coincides with the target of the other. Since C has an odd length, we have |E| > 1. For every {e, e′ } ∈ E consider the pair (a, b) ∈ MM(P, ν13 , Σ) which is the common endpoint of e and e′ in JΣ,1 , and let u{e,e′ } := uab . Choose {e, e′ } ∈ E such that u{e,e′ } is maximal in T , that is, u{e,e′ } 6< u{f,f ′ } in T for all {f, f ′ } ∈ E. Exchanging e and e′ if necessary we may assume that the target of e coincides with the source of e′ . Enumerate the vertices of the odd cycle C as (a1 , b1 ), (a2 , b2 ), . . . , (ak , bk ) in such a way that e = {(a1 , b1 ), (a2 , b2 )} and e′ = {(a2 , b2 ), (a3 , b3 )}. Let ui := uai bi , xi := xai bi , yi := yai bi , and zi := zai bi for each i ∈ {1, 2, . . . , k}. Thus u1 < u2 < u3 in T and u2 = u{e,e′ } . Let i be the largest index in {3, . . . , k} such that u2 < uj for all j ∈ {3, . . . , i} in T . If there is an index j ∈ {3, . . . , i} such that uj−1 < uj < uj+1 or uj−1 > uj > uj+1 in T (taking indices cyclically), then u{e,e′ } = u2 < uj in T , which contradicts our choice of {e, e′ } in E (since {{(aj−1 , bj−1 ), (aj , bj )}, {(aj , bj ), (aj+1 , bj+1 )}} was a better choice). Thus no such index j exists. Given that u2 < u3 in T , it follows that {uj−1 , uj+1 } < uj in T for each odd index j ∈ {3, . . . , i}, and that i is odd (because ui+1 < ui in T by the choice of i). Since u2 < ui and ui+1 < ui in T , the two nodes u2 and ui+1 are comparable in T . By our choice of i we have u2 6< ui+1 in T . (This is clear if i < k, and if i = k this follows from the fact that uk+1 = u1 < u2 in T .) It follows that ui+1 6 u2 in T . We claim that ui+1 < u2 in T . So suppose ui+1 = u2 . Observe that i 6= k (as otherwise ui+1 = u1 < u2 in T ) and i 6= k − 1 (since i and k are odd) in this case. Thus, 3 6 i 6 k − 2 and since the odd cycle C is induced, it follows that the two pairs (a1 , b1 ), (ai+1 , bi+1 ) do not form a 2-cycle. (For if they did, it would be a 2-cycle of type 1 since u1 < u2 = ui+1 in T , which would give a chord of C in JΣ,1 .) Applying Claim 27 on the pairs (a1 , b1 ), (a2 , b2 ) and (ai+1 , bi+1 ) we obtain a1 6 xi+1 in P . Using Claim 25 on (a1 , b1 ) and (a2 , b2 ) gives us x2 6 b1 in P . However, since u2 = ui+1 and (φ(x2 ), φ(y2 ), φ(z2 )) = (φ(xi+1 ), φ(yi+1 ), φ(zi+1 )) (given that α11 (a2 , b2 ) = α11 (ai+1 , bi+1 )), we deduce x2 = xi+1 , which implies a1 6 xi+1 = x2 6 b1 in P , a contradiction. Therefore, ui+1 < u2 in T , as claimed. In order to finish the proof, we consider separately the case i < k and i = k. First suppose that i < k, and thus i 6 k − 2. Since ui+1 < u2 < ui and u1 < u2 in T , using Claim 28 on the four pairs (ai+1 , bi+1 ), (ai , bi ), (a1 , b1 ), (a2 , b2 ) we obtain that {(ai+1 , bi+1 ), (a1 , b1 )} or {(ai+1 , bi+1 ), (a2 , b2 )} is an edge in JΣ,1 , showing that C has a chord, a contradiction. Next assume that i = k. Recall that {uj−1 , uj+1 } < uj in T for each odd index j ∈ {3, . . . , k}. It follows that uj−1 and uj+1 are comparable in T for each such index j. Using Observation 8 and k > 5 we deduce in particular that there exists an even index ℓ ∈ {4, . . . , k − 1} such that uℓ 6 uℓ′ for every even index ℓ′ ∈ {4, . . . , k − 1}. By the choice of ℓ we have uℓ < u3 in T (since uℓ 6 u4 < u3 ) and uℓ < uk in T (since uℓ 6 uk−1 < uk ). Note also that u1 < u2 < uℓ in T , since i = k. Applying Claim 29 on the five pairs (a1 , b1 ), (a2 , b2 ), (a3 , b3 ), (ak , bk ) and (aℓ , bℓ ), we then obtain that {(a1 , b1 ), (aℓ , bℓ )} or {(a2 , b2 ), (aℓ , bℓ )} is an edge in JΣ,1 , showing that C has a chord, a contradiction.  Now that the bipartiteness of JΣ,1 is established for each Σ ∈ Σ(ν13 ), to finish our proof of Lemma 22 (asserting that JΣ is 4-colorable), it remains to show that the two subgraphs of JΣ,2 induced by the two color classes in a 2-coloring of JΣ,1 are bipartite. Clearly, it is enough to show that every subgraph of JΣ,2 induced by an independent set of JΣ,1 is bipartite, which is exactly what we will do. (Recall that an independent set, also known as stable set, is a set of pairwise non-adjacent vertices.) To this aim we introduce a new definition: Given a pair (a, b) ∈ MM(P ) and a set {c, d} of two elements of P , we say that (a, b) is connected to {c, d} if a 6 c and d 6 b, or a 6 d and c 6 b in P . Note that if (a, b) is connected to {c, d}, then it is in exactly one of two possible ways (that is, either a 6 c and d 6 b, or a 6 d and c 6 b in P ). Thus two pairs (a, b), (a′ , b′ ) ∈ MM(P ) connected to {c, d} are either connected the same way, or in opposite ways. More generally, we can consider how a collection of pairs are connected to a certain set {c, d}, which we will need to do in what follows. Observation 31. Let Σ ∈ Σ(ν13 ). Let I be an independent set in JΣ,1 and let c, d be two distinct elements of P . Suppose that (a, b), (a′ , b′ ) ∈ I are two pairs that are connected to {c, d} in opposite ways. By 24 aT2 u2 aT1 bT2 aT3 s22 s32 bT3 u3 s2 bT1 s12 u1 s1 r Figure 11. Illustration for Claim 34 with j = 2 definition a 6 c 6 b′ and a′ 6 d 6 b, or a 6 d 6 b′ and a′ 6 c 6 b in P . Thus in both cases (a, b), (a′ , b′ ) is a 2-cycle, which must be of type 2 as (a, b) and (a′ , b′ ) are non-adjacent in JΣ,1 . In particular, (a, b) and (a′ , b′ ) are adjacent in JΣ,2 . Observation 32. Let Σ ∈ Σ(ν13 ). Let I be an independent set in JΣ,1 and let c, d be two distinct elements of P . Suppose that (a, b), (a′ , b′ ) ∈ I are adjacent in JΣ,2 and that the two relations a 6 b′ and a′ 6 b both hit {c, d}. Then (a, b), (a′ , b′ ) are connected to {c, d} in opposite ways. Claim 33. Let Σ ∈ Σ(ν13 ). Let I be an independent set in JΣ,1 and let c, d be two distinct elements of P . Suppose that C is an induced odd cycle in the subgraph of JΣ,2 induced by I. If four of the pairs composing C are connected to {c, d} then they all are connected to {c, d} the same way. Proof. Suppose that (a1 , b1 ), . . . , (a4 , b4 ) are four pairs from C that are connected to {c, d}. (Note that these pairs are not necessarily consecutive in C.) If three of these pairs are connected to {c, d} the same way and the fourth the other way, then by Observation 31 the fourth pair is adjacent to the first three in JΣ,2 , which is not possible since the odd cycle C is induced. It follows that if (a1 , b1 ), . . . , (a4 , b4 ) are not connected to {c, d} the same way, then without loss of generality (a1 , b1 ), (a2 , b2 ) are connected to {c, d} one way and (a3 , b3 ), (a4 , b4 ) the other. We then deduce from Observation 31 that (a1 , b1 ), (a3 , b3 ), (a2 , b2 ), (a4 , b4 ) is a cycle of length 4 in JΣ,2 , a contradiction to the properties of C. Therefore, all four pairs must be connected to {c, d} the same way.  Claim 34. Let Σ ∈ Σ(ν13 ) and let I be an independent set in JΣ,1 . Then the subgraph of JΣ,2 induced by I is bipartite. Proof. Arguing by contradiction, suppose there is an odd cycle in the subgraph of JΣ,2 induced by I, and let C be a shortest one. Enumerate the vertices of C as (a1 , b1 ), (a2 , b2 ), . . . , (ak , bk ) in order. Let ui := uai bi and si := ui ∧ ui+1 for each i ∈ {1, . . . , k} (cyclically). Recall that by the definition of JΣ,2 the pairs (ai , bi ) and (ai+1 , bi+1 ) form a 2-cycle of type 2 for each i, and thus ui k ui+1 in T , implying that si < {ui , ui+1 } in T . Let us start by pointing out the following consequence of Observation 32: If i ∈ {1, . . . , k} and {c, d} are such that the ui –ui+1 path in T includes an edge e of T for which the intersection of the two bags of its endpoints is {c, d}, then (ai , bi ) and (ai+1 , bi+1 ) are connected to {c, d} in opposite ways. This will be used a number of times in the proof. Let j ∈ {1, . . . , k} be such that sj is maximal in T among s1 , . . . , sk , that is, such that sj 6< si in T for each i ∈ {1, . . . , k}. Furthermore, if sj 6= ui then we let sij be the neighbor of sj on the sj –ui path in T , for each i ∈ {1, . . . , k}. (Let us remark that we only make use of the notion sij when sj 6= ui ). Thus in particular sj < {sjj , sj+1 } in T . See Figure 11 for an illustration of this definition. j We claim that B(sj ) ∩ B(ssj ) 6= B(sj ) ∩ B(stj ) for any s, t ∈ {1, . . . , k} with s < t such that sj < {ssj , stj } in T . Suppose to the contrary that B(sj ) ∩ B(ssj ) = B(sj ) ∩ B(stj ) =: {c, d}. It follows from our choice of index j that both the us−1 –us path and the us –us+1 path in T include the edge sj ssj (otherwise sj < ss in T ), and similarly that the ut –ut+1 path in T includes the edge sj stj (otherwise sj < st in T ). As a consequence we have that edge sj ssj is 25 included in the aTs−1 –bTs path (as it passes through us−1 and us ), the aTs –bTs−1 path, the aTs –bTs+1 path and the aTs+1 –bTs path, and similarly that edge sj stj is included in the aTt –bTt+1 path and the aTt+1 –bTt path. Therefore, the pairs (as−1 , bs−1 ), (as , bs ), (as+1 , bs+1 ), (at , bt ), and (at+1 , bt+1 ) are all connected to {c, d}. Furthermore, (as−1 , bs−1 ) and (as , bs ) are connected in opposite ways, and the same holds for (as , bs ) and (as+1 , bs+1 ), as well as for (at , bt ) and (at+1 , bt+1 ). There cannot be four distinct pairs among these five, because otherwise this would contradict Claim 33. Hence the only possibility is that k = 3 and s = t − 1, and thus s − 1 and t + 1 are the same indices cyclically. But then recall that the u1 –u2 path, the u2 –u3 path and the u3 –u1 path all have to use edge sj ssj or sj stj in T . As a consequence, the three relations a3 6 b1 , a1 6 b2 and a2 6 b3 all hit {c, d}. Hence two of them hit the same element, which implies ai 6 bi for some i ∈ {1, 2, 3}. With this contradiction, we have proved B(sj ) ∩ B(ssj ) 6= B(sj ) ∩ B(stj ). Since B(sj ) ∩ B(sij ) is a 2-element subset of B(sj ) for each i such that sj 6= ui , it directly follows that there are at most three indices i such that sj < sij in T (recall that this inequality holds for i = j and i = j + 1). This allows us to quickly dispense with the k > 5 case now: If uj−1 > sj in T (and hence sj−1 > sj j in T ), then let (s, t) := (j − 1, j + 1). Else, if uj+2 > sj in T , then let (s, t) := (j, j + 2). If neither of the two cases is true, we let (s, t) := (j, j + 1). We claim that in all three cases we obtain that • sj < {ssj , stj } in T , • the two indices s − 1 and t + 1 are not the same (cyclically), and • ss−1 = st+1 = p(sj ). j j The first two items are obvious, and the third one follows from our last observation and because sj 6= us−1 and sj 6= ut+1 (recall that us−1 k us and ut k ut+1 in T ). Then, the us−1 –us path and the ut –ut+1 path in T both include the edge p(sj )sj . It follows that (as−1 , bs−1 ) and (as , bs ) are connected to B(sj ) ∩ B(p(sj )) in opposite ways, and that the same holds for (at , bt ) and (at+1 , bt+1 ). Since these four pairs are distinct, this contradicts Claim 33 and concludes the case k > 5. It remains to consider the k = 3 case. Reordering the pairs of C if necessary we may assume j = 1 and B(s1 ) = {c, d, e} with B(s1 ) ∩ B(s11 ) = {c, d} and B(s1 ) ∩ B(s21 ) = {d, e}. The two relations a1 6 b2 and a2 6 b1 both hit {c, d} and {d, e}, and clearly they cannot hit the same element. Thus, one of the relations hits d, and the other c and e. Exploiting symmetry again, we may assume without loss of generality that a2 6 b1 hits d. (Indeed, if not then this can be achieved by reversing the ordering of the pairs of C.) Thus we have a 2 6 d 6 b1 in P , which then implies a 1 6 c 6 e 6 b2 in P by Observation 10. Now, the two relations a2 6 b3 and a3 6 b1 both hit B(s1 ) = {c, d, e}. (Here we use that s1 6< {s2 , s3 } in T .) Neither hit c or e since this would contradict a1 6 c 6 e 6 b2 in P . Hence both relations hit d, which implies a3 6 d 6 b3 in P , a contradiction.  This concludes the proof of Lemma 22, asserting that JΣ is 4-colorable for each Σ ∈ Σ(ν13 ). Now, for each Σ ∈ Σ(ν13 ) let ψ13,Σ be such a coloring. Then we define α13 as follows: For each Σ ∈ Σ(ν13 ) and pair (a, b) ∈ MM(P, ν13 , Σ), we let α13 (a, b) := ψ13,Σ (a, b). 3.12. Node ν14 and its function α14 : Dealing with strict alternating cycles of length at least 3. Recall that compared to MM(P, ν14 , Σ), pairs in MM(P, ν14 , Σ) have the additional property that they do not form a 2-cycle with another pair of MM(P, ν14 , Σ), thanks to function α13 . Therefore, strict alternating cycles in MM(P, ν14 , Σ) (Σ ∈ Σ(ν14 )) have length at least 3. We will now list (and prove) a number of properties satisfied by these alternating cycles. First we prove a claim that bears some similarity with Claim 12. Claim 35. Let Σ ∈ Σ(ν14 ) and suppose that {(ai , bi )}ki=1 is a strict alternating cycle in MM(P, ν14 , Σ). Let ui denote uai bi for each i ∈ {1, 2, . . . , k}. Then there is an index j ∈ {1, 2, . . . , k} such that uj 6 ui in T for each i ∈ {1, 2, . . . , k}. 26 Proof. We denote wai bi , pai bi , xai bi , yai bi , zai bi by wi , pi , xi , yi , zi respectively, for each i ∈ {1, 2, . . . , k}. We may assume α1 (ai , bi ) = left. Consider the nodes u1 , . . . , uk and let j ∈ {1, 2, . . . , k} be such that uj is minimal in T among these. We will show that uj 6 ui in T for each i ∈ {1, 2, . . . , k}. This can equivalently be rephrased as follows: Every element ui which is minimal in T among u1 , . . . , uk satisfies ui = uj (note that we could possibly have ui = uj for i 6= j). Arguing by contradiction, let us assume that there is an element minimal in T among u1 , . . . , uk which is distinct from uj . We start by showing that under this assumption u1 , . . . , uk are all pairwise incomparable in T (and thus are in particular all distinct). Once this is established, we will then be able to derive the desired contradiction. Of course, to prove that u1 , . . . , uk are pairwise incomparable in T it is enough to show that ui k uj in T for each i ∈ {1, 2, . . . , k} with i 6= j, since uj was chosen as an arbitrary minimal element in T among u1 , . . . , uk . Assume not, that is, that there is an index i ∈ {1, 2, . . . , k} with i 6= j such that uj 6 ui in T . We may choose i in such a way that we additionally have ui−1 k uj or ui+1 k uj in T . As the arguments for the two cases are analogous we consider only the case ui−1 k uj in T . We have aTi−1 6> uj since ui−1 k uj in T , and we also have uj 6 ui < bTi in T . It follows that the path from aTi−1 to bTi in T goes through the edge pj uj . Thus, the relation ai−1 6 bi hits B(pj ) ∩ B(uj ) = {xj , yj }. But then ai−1 6 xj 6 bi or ai−1 6 yj 6 bi in P , which implies aj 6 xj 6 bi or ai−1 6 yj 6 bj . Since we have i 6= j + 1 (as ui−1 k uj in T ) and j 6= i, this contradicts the assumption that our alternating cycle (a1 , b1 ), (a2 , b2 ), . . . , (ak , bk ) is strict. We conclude that u1 , . . . , uk are all pairwise incomparable in T , as claimed. Let si := ui ∧ ui+1 for each i ∈ {1, 2, . . . , k} (indices are taken cyclically, as always). Note that the path from aTi to bTi+1 in T has to go through si . Choose i ∈ {1, . . . , k} such that si is maximal among s1 , . . . , sk in T . The nodes si−1 and si are comparable in T , since si−1 6 ui and si 6 ui in T . Thus we have si−1 6 si in T . Similarly, si+1 6 si in T . Let us first look at the case si−1 = si . This implies si 6 {ui−1 , ui , ui+1 } in T . Now the aTi−1 –bTi path, the aTi –bTi+1 path, and the aTi+1 –bTi+2 path in T all go through si in T . This means that the relations ai−1 6 bi , ai 6 bi+1 and ai+1 6 bi+2 all hit B(si ). Clearly, no two of them can hit the same element (recall that k > 3 and that our alternating cycle is strict), and hence each element of B(si ) is hit by exactly one of these three relations. On the other hand, the three paths from r to bTi−1 , bTi and bTi+1 in T all go through p(si ) and si , implying that the relations a0 6 bi−1 , a0 6 bi , and a0 6 bi+1 all hit B(p(si )) ∩ B(si ). In particular, some element in B(si ) is hit by at least two of these three relations. But with the observations made before, it follows that some element in {ai−1 , ai , ai+1 } is below two elements of {bi−1 , bi , bi+1 } in P , which is not possible in a strict alternating cycle. Therefore, si−1 6= si . Thus we have si−1 < si in T , and with a similar argument one also deduces that si+1 < si in T . To conclude the proof, consider the aTi−1 –bTi path, the aTi+1 –bTi+2 path, and the r–bTi+1 path in T . They all go through the edge p(si )si of T , and hence the corresponding relations in P all hit B(p(si )) ∩ B(si ). Therefore, two of these relations hit the same element in that set, which again contradicts the fact that our alternating cycle is strict.  By Claim 35 we are in a situation similar to that first encountered in Section 3.5, namely for each Σ ∈ Σ(ν14 ) each alternating cycle in MM(P, ν14 , Σ) can be written as {(ai , bi )}ki=1 in such a way that ua1 b1 6 uai bi in T for each i ∈ {1, . . . , k}. We may further assume that the pair (a1 , b1 ) is such that bT1 is to the right of bTi in T if aT1 is to the left of bT1 in T , and to the left of bTi otherwise, for each i ∈ {2, . . . , k} such that ua1 b1 = uai bi . As before the pair (a1 , b1 ) is uniquely defined, and we call it the root of the alternating cycle. Our next claim mirrors Claim 14 from Section 3.5. Claim 36. Let Σ ∈ Σ(ν14 ) and suppose that {(ai , bi )}ki=1 is a strict alternating cycle in MM(P, ν14 , Σ) with root (a1 , b1 ). Let ui , wi denote uai bi , wai bi respectively, for each i ∈ {1, 2, . . . , k}. Then u1 < w1 6 ui in T for each i ∈ {2, . . . , k}. Proof. We denote pai bi , xai bi , yai bi , zai bi by pi , xi , yi , zi respectively, for each i ∈ {1, 2, . . . , k}. We may assume α1 (ai , bi ) = left for each i ∈ {1, 2, . . . , k}. First we will show that u1 < w1 6 uk in T . To do so suppose first that u1 = uk . Then w1 66 aTk in T , as otherwise we would have bTk to the right of bT1 in T (Observation 9), which contradicts the choice of (a1 , b1 ) as the root of the strict alternating cycle. In particular, the path from aTk to bT1 in T goes through u1 . Hence the relation ak 6 b1 hits B(u1 ) = {x1 , y1 , z1 }; let q ∈ B(u1 ) be such that ak 6 q 6 b1 in P . Clearly, q ∈ {y1 , z1 }. Given that u1 = uk and (φ(x1 ), φ(y1 ), φ(z1 )) = (φ(xk ), φ(yk ), φ(zk )) 27 (since α11 (a1 , b1 ) = α11 (ak , bk )), we obviously have x1 = xk , y1 = yk , and z1 = zk . If q = y1 = yk then we directly obtain q 6 bk in P . If q = z1 = zk then we also deduce q 6 bk in P , because α12 (a1 , b1 ) = α12 (ak , bk ), and thus in particular zk 6 bk in P since z1 6 b1 . Hence in both cases q 6 bk in P . This implies ak 6 q 6 bk in P , a contradiction. Therefore, u1 6= uk , and u1 < uk in T . Let w′ be the neighbor of u1 on the u1 –uk path in T . In order to show u1 < w1 6 uk in T , it remains to prove w′ = w1 . Suppose to the contrary that w′ 6= w1 . Then the aTk –bT1 path and the r–bTk path in T both go through u1 and w′ . Hence the relations ak 6 b1 and a0 6 bk both hit B(u1 ) ∩ B(w′ ) {x1 , y1 , z1 }. Clearly, they cannot hit the same element. None of the two relations hit x1 , as otherwise a1 6 x1 6 b1 or a1 6 x1 6 bk in P (which is not possible since k > 3 and the alternating cycle is strict). We conclude that B(u1 ) ∩ B(w′ ) = {y1 , z1 }. Since the relation a0 6 bk also hits B(u1 ) ∩ B(p1 ) = {x1 , y1 }, and thus hits y1 , it follows that a0 6 y1 6 bk and ak 6 z1 6 b1 in P . Now let i ∈ {1, . . . , k − 1} be maximal such that w′ 66 ui in T . Note that there is such an index since w′ 66 u1 in T . If w′ 6 aTi in T , then the path from aTi to ui in T goes through the edge u1 w′ . If, on the other hand, w′ k aTi in T , then the path from aTi to bTi+1 in T goes through the edge u1 w′ . Thus at least one of ai 6 xi and ai 6 bi+1 hits B(u1 ) ∩ B(w′ ) = {y1 , z1 }. Hence ai 6 y1 or ai 6 z1 in P . However, since {y1 , z1 } 6 b1 in P , this implies ai 6 b1 in both cases. Given that i < k, this contradicts the fact that the alternating cycle is strict. Therefore, we must have w′ = w1 , and u1 < w1 6 uk in T , as claimed. So far we know that w1 6 ui in T for i = k, and it remains to show it for each i ∈ {2, . . . , k − 1}. Arguing by contradiction, assume that this does not hold, and let i ∈ {2, . . . , k − 1} be maximal such that w1 66 ui in T . By our choice it holds that w1 6 ui+1 in T , even in the case i = k − 1. First suppose that ui = u1 . Then w1 66 aTi in T , because otherwise bTi would be to the right of bT1 in T (Observation 9), contradicting the fact that (a1 , b1 ) is the root of the strict alternating cycle. But then, the path from aTi to bTi+1 goes through the edge u1 w1 (since w1 6 ui+1 in T ). Thus the relation ai 6 bi+1 hits in particular B(u1 ); let q ∈ B(u1 ) be such that ai 6 q 6 bi+1 in P . Given that u1 = ui we deduce x1 = xi , y1 = yi , and z1 = zi (using α11 (a1 , b1 ) = α11 (ai , bi )), and hence a1 6 q in P (using α12 (a1 , b1 ) = α12 (ai , bi )), exactly as in the beginning of the proof. This implies a1 6 q 6 bi+1 in P , and as i + 1 > 3 this contradicts once again the fact that the alternating cycle is strict. Therefore, u1 6= ui , and u1 < ui in T . Let w′ be the neighbor of u1 on the u1 –ui path in T . Note that w′ 6= w1 . The aTi –bTi+1 path and the r–bTi path both go through the edge u1 w′ . Thus the relations ai 6 bi+1 and a0 6 bi both hit B(u1 ) ∩ B(w′ ) {x1 , y1 , z1 }. Clearly, they cannot hit the same element. Since ai 6 x1 6 bi+1 in P would imply a1 6 bi+1 (which is not possible since i + 1 > 3) while ai 6 y1 6 bi+1 would imply ai 6 b1 (which cannot be since i < k), we deduce ai 6 z1 6 bi+1 in P , and a 0 6 q 6 bi in P , where q is the element in {x1 , y1 } such that B(u1 ) ∩ B(w′ ) = {q, z1 }. We distinguish two cases, depending whether q = x1 or q = y1 . First suppose that q = x1 . Since a0 6 q = x1 6 bi in P , this implies a1 6 x1 6 bi in P , and hence i = 2 (otherwise, the alternating cycle would not be strict). Furthermore, given that a0 6 x1 in P and α12 (a1 , b1 ) = α12 (a2 , b2 ), we have a0 6 x2 in P as well. The r–u2 path in T includes the edge u1 w′ since w′ 6 u2 in T . Using that x2 ∈ B(u2 ), we deduce that the relation a0 6 x2 in P hits B(u1 ) ∩ B(w′ ) = {x1 , z1 }. In particular, at least one of x1 6 x2 and z1 6 x2 holds in P . Before considering each of these two possibilities, let us observe that the aT2 –u1 path in T includes the edge u2 p2 . It follows that the relation a2 6 z1 hits B(u2 ) ∩ B(p2 ) = {x2 , y2 }. Clearly, it cannot hit y2 (otherwise a2 6 y2 6 b2 ), and hence x2 6 z1 in P . Now, if z1 6 x2 in P then x2 = z1 . However, we also know that φ(x2 ) = φ(x1 ) 6= φ(z1 ), since α11 (a1 , b1 ) = α11 (a2 , b2 ), which is a contradiction. On the other hand, if x1 6 x2 in P then a1 6 x1 6 x2 6 z1 6 bi+1 = b3 in P , which contradicts the fact the alternating cycle is strict. This concludes the case where q = x1 . Next, assume q = y1 . Let j ∈ {1, . . . , i − 1} be maximal such that w′ 66 uj in T . (Note that there is such an index j since w′ 66 u1 in T .) If w′ 6 aTj in T then the path from aTj to uj in T goes through the edge u1 w′ . If, on the other hand, w′ k aTj in T , then the path from aTj to bTj+1 in T goes through the edge u1 w′ since w′ 6 uj+1 < bTj+1 in T . Hence at least one of the two relations aj 6 xj and aj 6 bj+1 hits {q, z1 } = {y1 , z1 }. It follows that aj 6 y1 or aj 6 z1 in P . The first inequality implies aj 6 y1 6 b1 in P , a contradiction since j 6= k. The second inequality implies aj 6 z1 6 bi+1 in P , which is not possible since j 6= i. This concludes the proof.  28 Claim 37. Let Σ ∈ Σ(ν14 ) and suppose that {(ai , bi )}ki=1 is a strict alternating cycle in MM(P, ν14 , Σ) with root (a1 , b1 ). Let ui denote uai bi for each i ∈ {1, 2, . . . , k}. Then the u1 –bT1 path in T avoids u2 . Proof. We denote wai bi , pai bi , xai bi , yai bi , by wi , pi , xi , yi respectively, for each i ∈ {1, 2, . . . , k}. We may assume α1 (ai , bi ) = left. Arguing by contradiction, suppose that u1 6 u2 < bT1 in T . By Claim 36 we know u1 < w1 6 u2 < bT2 in T . The aT1 –bT2 path in T goes through the edge p2 u2 . Hence the relation a1 6 b2 hits B(p2 ) ∩ B(u2 ) = {x2 , y2 }. Clearly, it cannot hit x2 because otherwise a2 6 x2 6 b2 in P . Therefore, a1 6 y2 6 b2 in P . Now consider the path connecting r to bT1 in T . This path also includes the edge p2 u2 . Thus the relation a0 6 b1 hits {x2 , y2 }. If it hits x2 , then we obtain a2 6 x2 6 b1 in P , which contradicts the fact that the alternating cycle is strict (recall that k > 3). If it hits y2 , then we deduce a1 6 y2 6 b1 in P , again a contradiction.  Let Σ ∈ Σ(ν14 ) and suppose that {(ai , bi )}ki=1 is a strict alternating cycle in MM(P, ν14 , Σ) with root (a1 , b1 ). In what follows we will need to consider the nodes qi := ui ∧ bT1 of T where i ∈ {1, 2, . . . , k}. Observe that u1 < w 1 6 qi 6 u i in T for each i ∈ {2, 3, . . . , k} by Claim 36, and q 2 < u2 in T by Claim 37. Claim 38. Let Σ ∈ Σ(ν14 ) and suppose that {(ai , bi )}ki=1 is a strict alternating cycle in MM(P, ν14 , Σ) with root (a1 , b1 ). Let ui denote uai bi and let qi := ui ∧ bT1 , for each i ∈ {1, 2, . . . , k}. Then (i) ui 66 q2 in T for each i ∈ {3, 4, . . . , k}, and (ii) u1 < q2 6 q3 < bT1 in T . Proof. We denote wai bi , pai bi , xai bi , yai bi , by wi , pi , xi , yi respectively, for each i ∈ {1, 2, . . . , k}. We may assume α1 (ai , bi ) = left. To prove (i) we argue by contradiction: Suppose ui 6 q2 in T for some i ∈ {3, 4, . . . , k}. Since q2 < bT1 in T , and u1 < w1 6 ui by Claim 36, it follows that u1 < w1 6 ui 6 q2 < bT1 in T . In particular, the path connecting aT1 to bT2 in T goes through the edge pi ui . Hence the relation a1 6 b2 hits B(pi ) ∩ B(ui ) = {xi , yi }. If it hits xi then ai 6 xi 6 b2 in P , while if it hits yi then a1 6 yi 6 bi in P . In both cases it contradicts the fact that the alternating cycle is strict. Let us now prove (ii). Using Claim 36 we already deduce that u1 < {q2 , q3 } < bT1 in T . Thus, it remains to show q2 6 q3 in T . Arguing by contradiction, suppose q3 < q2 in T (note that q2 and q3 are comparable in T ). Let i be the largest index in {3, 4, . . . , k} such that qi < q2 in T . If i < k then qi < q2 6 qi+1 6 ui+1 < bTi+1 in T . If i = k then clearly qi < q2 < bT1 in T . Thus in both cases qi < q2 < bTi+1 (22) q2 66 aTi (23) in T (taking indices cyclically). Observe also that aTi bTi in T . Indeed, if q2 6 in T then q2 6 as well, since otherwise ui < q2 in T , contradicting (i). However, this implies q2 6 ui in T , and hence q2 6 qi since q2 < bT1 , a contradiction. Now consider the edge p(q2 )q2 in T and let B(p(q2 )) ∩ B(q2 ) = {c, d}. In the following, we aim to show that the relevant part of T essentially looks like in Figure 12, and consequently that the relations ai 6 bi+1 and a2 6 b3 have to hit {c, d}. From this observation we will obtain our final contradiction. Using (23) and that q2 6 bTi+1 in T (see (22)), we deduce that the path from aTi to bTi+1 in T goes through this edge. Thus the relation ai 6 bi+1 hits {c, d}. Without loss of generality ai 6 c 6 bi+1 (24) in P . To see that a2 6 b3 also hits {c, d} we first show that q2 66 bT3 in T . For this suppose q2 6 bT3 in T . Then q2 and u3 are comparable in T , and thus q2 < u3 in T by (i). Since q2 < bT1 in T , it follows that q2 6 u3 ∧ bT1 = q3 in T , contradicting our assumption that q3 < q2 in T. 29 aT3 T T T bT3 a2 b2 a4 aT1 u4 u2 u3 cd bT4 aT2 aT1 T aT4 b4 u3 u2 bT1 q4 = qi+1 q2 bT2 aT3 bT3 w w′=w′′ u4 bT1 q4 = qi+1 q2 = q 3 = qi q3 = q i u1 u1 Figure 12. Left: Situation in Claim 38 with i = 3 (under the assumption that q3 < q2 in T ). Right: Possible situation in Claim 39 with i = 3 (under the assumptions that q2 = q3 and w 6= w′ ). So we have q2 66 bT3 , and since q3 < q2 6 u2 < aT2 in T , we deduce that the path connecting aT2 to bT3 in T also includes the edge p(q2 )q2 . Thus the relation a2 6 b3 indeed hits {c, d}. It cannot hit c because otherwise a2 6 c 6 bi+1 (by (24)), which is not possible since i 6= 2. Hence we have (25) a 2 6 d 6 b3 in P . Now, the relation a0 6 b2 clearly hits {c, d} as well, but this is not possible as this implies ai 6 c 6 b2 (using (24)) or a2 6 d 6 b2 in P (using (25)), a contradiction in both cases. This concludes the proof of (ii).  The following claim is a strengthening of Claim 38 (ii). Claim 39. Let Σ ∈ Σ(ν14 ) and suppose that {(ai , bi )}ki=1 is a strict alternating cycle in MM(P, ν14 , Σ) with root (a1 , b1 ). Let ui denote uai bi and let qi := ui ∧ bT1 , for each i ∈ {1, 2, . . . , k}. Then u1 < q2 < q3 < bT1 in T. Proof. Using Claim 38 (ii) we only need to show that q2 6= q3 . Suppose to the contrary that we have q2 = q3 . Recall that q2 < u2 in T , by Claim 37. By Claim 38 (i) we cannot have u3 = q3 since q2 = q3 . Hence we also have q3 < u3 in T . Now, let w be the neighbor of q2 on the path from q2 to u2 in T , and let w′ be the neighbor of q3 on the path from q3 = q2 to u3 in T . Using that q2 < u2 and q3 < u3 in T we deduce that q2 < w 6 u2 and q3 < w ′ 6 u3 (26) ′ in T . Let us first suppose that w = w . Let i be the largest index in {3, 4, . . . , k} such that w 6 ui in T . We claim that w 66 bTi+1 in T (taking indices cyclically, as always). If i < k this is because w 6 bTi+1 would imply w 66 aTi+1 (since w 66 ui+1 in T ), and thus ui+1 6 q2 in T , contradicting (i) of Claim 38. If i = k this is because w 6 bT1 together with w 6 u2 (by (26)) would imply w 6 u2 ∧ bT1 = q2 in T , a contradiction. Now, since w 66 bTi+1 in T we deduce that the path from aTi to bTi+1 includes the edge q2 w. Let B(w) ∩ B(q2 ) = {c, d}. Then the relations ai 6 bi+1 and a1 6 b2 both hit {c, d}, but not the same element (as otherwise ai 6 b2 in P , which cannot be since i 6= 1). Say we have ai 6 c 6 bi+1 and a 1 6 d 6 b2 in P . Observe that the path from r to bTi in T goes through the edge q2 w as well, and hence a0 6 bi also hits {c, d}. Thus c 6 bi or d 6 bi in P . In the first case we obtain ai 6 bi and in the second a1 6 bi , a contradiction in each case. This closes the case w = w′ . Finally, assume w 6= w′ . Let B(q2 ) = {c, d, e}. Let i be the largest index in {3, 4, . . . , k} such that qi = q2 . By Claim 38 (i) we know that qi 6= ui (in particular ui 6< bT1 in T ), and thus q2 = qi < ui in T . Let w′′ be the neighbor of q2 on the path from q2 to ui in T . Note that we must have q2 < w′′ 6 ui and w′′ k bT1 (27) in T . Next we show that w′′ 66 bTi+1 in T . Suppose that this is not true, and let us consider the case i < k first. Then w′′ 6 bTi+1 would imply w′′ 6 aTi+1 as otherwise ui+1 6 q2 in T , contradicting (i) of Claim 38. Moreover, this yields w′′ 6 ui+1 30 u2 m(q2 ) bT1 ce de n(q2 ) cd q2 p(q2 ) u1 Figure 13. Illustration of Claim 40. in T . However, combined with the fact that q2 6 bT1 and w′′ 66 bT1 in T (by (27)) this implies qi+1 = q2 , contradicting the choice of i. If i = k then w′′ 6 bT1 together with w′′ 6 ui (see (27)) would imply w′′ 6 qi , again a contradiction. So we indeed have w′′ 66 bTi+1 in T (for an example illustrating this situation with i = 3 see Figure 12 on the right), and from this it follows that the aTi –bTi+1 path in T includes the node q2 . Observe that so does the aT1 –bT2 path (because u1 < w1 6 q2 < bT2 in T by Claim 36) and the aT2 –bT3 path (because w 6= w′ ). Hence the three relations a1 6 b2 , a2 6 b3 and ai 6 bi+1 all hit B(q2 ) = {c, d, e}. Clearly, no element in B(q2 ) is hit by two of these. In other words, each element of B(q2 ) is greater or equal to a1 , a2 , or ai in P . Furthermore, the paths from r to bT1 , bT2 and bT3 in T all include the edge p(q2 )q2 . Hence two of the three relations a0 6 b1 , a0 6 b2 , a0 6 b3 hit the same element in B(p(q2 )) ∩ B(q2 ) {c, d, e}. It follows that one element of the set {a1 , a2 , ai } is below two different elements of {b1 , b2 , b3 } in P , which contradicts the assumption that the alternating cycle is strict. This concludes the proof.  Let Σ ∈ Σ(ν14 ) and suppose that {(ai , bi )}ki=1 is a strict alternating cycle in MM(P, ν14 , Σ) with root (a1 , b1 ). Let ui denote uai bi for each i ∈ {1, 2, . . . , k}, and let q2 := u2 ∧ bT1 . In the following claims we will need to consider three specific neighbors of the node q2 in T , namely, the neighbors of q2 on the q2 –u1 path, the q2 –u2 path, and the q2 –bT1 path in T . Let us denote these nodes by p(q2 ), m(q2 ) and n(q2 ), respectively. By Claims 36 and 37, p(q2 ), m(q2 ) and n(q2 ) are well defined and distinct. The following claim is illustrated in Figure 13. Claim 40. Let Σ ∈ Σ(ν14 ) and suppose that {(ai , bi )}ki=1 is a strict alternating cycle in MM(P, ν14 , Σ) with root (a1 , b1 ). Let ui denote uai bi and let qi := ui ∧ bT1 , for each i ∈ {1, 2, . . . , k}. Then the elements of B(q2 ) can be written as B(q2 ) = {c, d, e} in such a way that (i) B(q2 ) ∩ B(p(q2 )) = {c, d}; (ii) B(q2 ) ∩ B(m(q2 )) = {c, e}; (iii) B(q2 ) ∩ B(n(q2 )) = {d, e}; and so that in P we have a 1 6 c 6 b2 ; (iv) a 0 6 d 6 b1 ; a 2 6 e 6 b3 ; (v) c d; e d; c k e; (vi) a2 c; a2 d. Proof. By Claims 36–39 we know that u1 < w1 6 q2 < u2 < {aT2 , bT2 } and q2 < q3 < {bT1 , aT3 , bT3 } in T . Thus the aT1 –bT2 path in T goes through the nodes p(q2 ), q2 , and m(q2 ); the r–bT1 path goes through p(q2 ), q2 , and n(q2 ), and the aT2 –bT3 path goes through m(q2 ), q2 , and n(q2 ). It follows that the corresponding three relations a1 6 b2 , a0 6 b1 and a2 6 b3 in P hit respectively the two sets B(q2 ) ∩ B(p(q2 )) and B(q2 ) ∩ B(m(q2 )); the two sets B(q2 ) ∩ B(p(q2 )) and B(q2 ) ∩ B(n(q2 )), and the two sets B(q2 ) ∩ B(m(q2 )) and B(q2 ) ∩ B(n(q2 )). Clearly, no element of B(q2 ) is hit by two of these three relations. It follows that the elements of B(q2 ) can be written as B(q2 ) = {c, d, e} in such a way that properties (i)-(iv) hold. The remaining two properties (v) and (vi) are immediate consequences of these.  For each Σ ∈ Σ(ν14 ) we define a corresponding directed graph K̂Σ on the set MM(P, ν14 , Σ) similarly as in Section 3.8: Given two distinct pairs (a1 , b1 ), (a2 , b2 ) ∈ MM(P, ν14 , Σ), there is an arc from (a1 , b1 ) to (a2 , b2 ) in K̂Σ if and only if there is a strict alternating cycle {(a′i , b′i )}ki=1 in MM(P, ν14 , Σ) with root (a′1 , b′1 ) which is such that (a′1 , b′1 ) = (a1 , b1 ) and (a′2 , b′2 ) = (a2 , b2 ). In the latter case, we say that the arc f is induced by the strict alternating cycle {(a′i , b′i )}ki=1 . 31 Note that there could possibly be different strict alternating cycles inducing the same arc in K̂Σ . Observe also that if {(ai , bi )}ki=1 is a strict alternating cycle inducing an arc in K̂Σ then (a1 , b1 ) is always the root of the cycle (by the definition of ‘inducing’). For each arc f = ((a1 , b1 ), (a2 , b2 )) of K̂Σ , define the corresponding three nodes of T : u− (f ) := ua1 b1 ; u+ (f ) := ua2 b2 ; q(f ) := ua2 b2 ∧ bT1 . Observe that u− (f ) < wa1 b1 6 q(f ) < u+ (f ) in T by Claims 36 and 37. This will be used repeatedly in what follows. Claim 41. For each Σ ∈ Σ(ν14 ), any two arcs f, g in K̂Σ sharing the same source satisfy q(f ) = q(g). Proof. Assume to the contrary that q(f ) 6= q(g). Let (a1 , b1 ) ∈ MM(P, ν14 , Σ) denote the source of the two arcs f and g. By definition q(f ) < bT1 and q(g) < bT1 in T . Thus in particular q(f ) and q(g) are comparable in T , say without loss of generality q(f ) < q(g). Hence we have ua1 b1 < wa1 b1 6 q(f ) < q(g) < bT1 in T . Let (a2 , b2 ), (a′2 , b′2 ) ∈ MM(P, ν14 , Σ) denote the targets of arcs f and g, respectively. Let (a3 , b3 ), . . . , (ak , bk ) ∈ MM(P, ν14 , Σ) be such that {(ai , bi )}ki=1 is a strict alternating cycle inducing f . Write the elements of B(q(f )) as B(q(f )) = {c, d, e} as in Claim 40 when applied to the latter cycle. Then T the paths from aT1 to b′T 2 and from r to b1 in T both include the three nodes p(q(f )), q(f ), and n(q(f )). ′ Hence, the two relations a1 6 b2 and a0 6 b1 in P both hit the two sets B(q(f )) ∩ B(p(q(f ))) = {c, d} and B(q(f )) ∩ B(n(q(f ))) = {d, e}. On the other hand, each of c, d, e is clearly hit by at most one of these two relations. It follows that one relation hits d and the other hits both c and e, implying c 6 e in P by Observation 10. However, this contradicts c k e in P (cf. property (v) of Claim 40).  The following claim is similar to the previous one. Claim 42. For each Σ ∈ Σ(ν14 ), any two arcs f, g in K̂Σ sharing the same target satisfy q(f ) = q(g). Proof. Assume to the contrary that q(f ) 6= q(g). Let (a2 , b2 ) ∈ MM(P, ν14 , Σ) denote the common target of the two arcs f and g. We have q(f ) 6 ua2 b2 and q(g) 6 ua2 b2 in T . Thus q(f ) and q(g) are comparable in T , say q(f ) < q(g) in T . Applying Claim 40 on a strict alternating cycle inducing f , we see that there exists an element e ∈ B(q(f )) such that a2 6 e in P . Since q(f ) < q(g) 6 ua2 b2 < aT2 in T , the path from q(f ) to aT2 in T goes through p(q(g)) and q(g). Thus the relation a2 6 e hits B(q(g)) ∩ B(p(q(g))), and hence a2 6 s in P for some s ∈ B(q(g)) ∩ B(p(q(g))). However, applying Claim 40 on a strict alternating cycle inducing g this time, we deduce that a2 t in P for each t ∈ B(q(g)) ∩ B(p(q(g))) (cf. property (vi)), and therefore in particular a2 s in P , a contradiction.  Claim 43. For each Σ ∈ Σ(ν14 ) and any two arcs f, g in K̂Σ , we neither have q(f ) < q(g) < u+ (f ) 6 u+ (g) nor q(f ) < q(g) < u+ (g) 6 u+ (f ) in T . Proof. Let (a2 , b2 ) and (a′2 , b′2 ) denote the targets of f and g, respectively. Arguing by contradiction, assume that at least one of the two inequalities holds. Then we have q(f ) 6 p(q(g)) < q(g) < m(q(g)) 6 {ua2 b2 , ua′2 b′2 } (28) in T . Now consider a strict alternating cycle inducing g and write the elements of B(q(g)) as B(q(g)) = {c, d, e} as in Claim 40 when applied to the latter cycle. See Figure 14 for an illustration. By this claim we have c 6 b′2 and a′2 6 e and e d (29) in P . Applying Claim 40 on a strict alternating cycle inducing f , we also deduce that there exists s ∈ B(q(f )) such that a2 6 s in P . 32 u+ (f ) u+ (g) ce de cd q(g) q(f ) Figure 14. Illustration of the proof of Claim 43. Given that by (28) the path from aT2 to q(f ) goes first through ua2 b2 and then through the three nodes m(q(g)), q(g) and p(q(g)) in T , it follows that the relation a2 6 s hits both B(q(g)) ∩ B(m(q(g))) = {c, e} and B(q(g)) ∩ B(p(q(g))) = {c, d}. If it did not hit c, then it would hit both d and e, and we would have e 6 d in P by Observation 10, which is not possible by (29). Thus a2 6 s hits c, that is, a2 6 c 6 s in P . Together with (29) this implies a2 6 c 6 b′2 in P , and we also deduce (a2 , b2 ) 6= (a′2 , b′2 ). Now, the path from r to bT2 in T goes through q(g) and m(q(g)), and thus a0 6 b2 hits {c, e}. It cannot hit c, as otherwise a2 6 c 6 b2 in P . Hence a0 6 b2 hits e, and we have a0 6 e 6 b2 in P , which by (29) implies a′2 6 e 6 b2 . It follows that (a2 , b2 ), (a′2 , b′2 ) is an alternating cycle of length 2, which is a contradiction since there is no such cycle in MM(P, ν14 , Σ).  For the next claim let us recall that given a pair (a, b) ∈ MM(P, ν14 , Σ), the elements of B(uab ) are labeled with xab , yab , zab , and it holds that B(uab ) ∩ B(p(uab )) = {xab , yab }, a 6 xab 66 b, and a 66 yab 6 b in P . Claim 44. For each Σ ∈ Σ(ν14 ), no two arcs f, g in K̂Σ satisfy u− (f ) 6 q(g) < u+ (g) 6 q(f ). in T . Proof. Assume to the contrary that the inequality holds. Let {(ai , bi )}ki=1 be a strict alternating cycle inducing f and let {(a′i , b′i )}ℓi=1 be one inducing g. Let ui := uai bi , qi := uai bi ∧bT1 , xi := xai bi , yi := yai bi , ′ ′ and zi := zai bi for each i ∈ {1, . . . , k}, and let u′i := ua′i b′i , qi′ := ua′i b′i ∧ b′T 1 , xi := xa′i b′i , yi := ya′i b′i , and ′ − + ′ ′ ′ ′ zi := zai bi for each i ∈ {1, . . . , ℓ}. Thus u (f ) = u1 , q(f ) = q2 , u (g) = u2 , q(g) = q2 , and u1 6 q2′ < u′2 6 q2 < {bT1 , bT2 } (30) in T by our assumption. See Figure 15 for an illustration of the situation. The aT1 –bT2 path and the r–bT1 path both go through p(u′2 ) and u′2 in T . Thus the two relations a1 6 b2 and a0 6 b1 both hit B(p(u′2 )) ∩ B(u′2 ) = {x′2 , y2′ }, and clearly neither of x′2 , y2′ is hit by both relations. We cannot have a1 6 y2′ 6 b2 and a0 6 x′2 6 b1 in P , because otherwise we would have a1 6 y2′ 6 b′2 and a′2 6 x′2 6 b1 in P , implying that (a1 , b1 ), (a′2 , b′2 ) is an alternating cycle of length 2 in MM(P, ν14 , Σ), a contradiction. Hence we have a1 6 x′2 6 b2 and a0 6 y2′ 6 b1 (31) in P . Let us denote the elements in B(q2′ ) as B(q2′ ) = {c, d, e} as in Claim 40 when applied to the strict alternating cycle {(a′i , b′i )}ℓi=1 . Then we have a0 6 d 6 b′1 in P , as well as a′2 6 e 6 b′3 and a′1 6 c 6 b′2 . Given that the relation a′2 6 e hits B(p(u′2 )) ∩ B(u′2 ) = {x′2 , y2′ }, and that it clearly cannot hit y2′ because 33 u2 bT1 q2 u′2 x′2 y2′ ce aT 1 q2′ cd u1 Figure 15. Illustration of the proof of Claim 44. y2′ 6 b′2 in P , we have a′2 6 x′2 6 e in P . Similarly, c 6 b′2 hits {x′2 , y2′ } as well and cannot hit x′2 because a′2 6 x′2 in P , hence c 6 y2′ 6 b′2 in P . Summarizing, we have a′2 6 x′2 6 e 6 b′3 , (32) a′1 (33) 6c6 y2′ a0 6 6 b′2 , d 6 b′1 (34) in P . Recall that u1 6 q2′ in T by (30). We split the rest of the argument into two cases and start with the case that u1 < q2′ in T . Then the path from aT1 to u′2 goes through p(q2′ ) and q2′ . It follows that the relation a1 6 x′2 hits B(p(q2′ )) ∩ B(q2′ ) = {c, d}. If it hits c then c 6 x′2 in P , which implies a′1 6 c 6 x′2 6 e 6 b′3 by (33) and (32), a contradiction to the assumption that we deal with a strict alternating cycle of length at least 3. Hence, we have a1 6 d 6 x′2 . However, using (34) we obtain a1 6 d 6 b′1 in P , and since a′1 6 c 6 y2′ 6 b1 in P (by combining (33) and (31)) this implies that (a1 , b1 ) 6= (a′1 , b′1 ) and therefore that (a1 , b1 ), (a′1 , b′1 ) is an alternating cycle of length 2 in MM(P, ν14 , Σ), a contradiction. It remains to consider the case that u1 = q2′ in T . Then the path from aT1 to u′2 in T goes through ′ q2 and m(q2′ ). Thus a1 6 x′2 hits B(q2′ ) ∩ B(m(q2′ )) = {c, e}. The relation a1 6 x′2 cannot hit c, for the same reason as in the previous paragraph. Hence a1 6 e 6 x′2 in P , implying that x′2 = e by (32). Now, observe that e ∈ B(u1 ) since u1 = q2′ . Given that e 6∈ B(q2′ ) ∩ B(p(q2′ )) = {c, d} = B(u1 ) ∩ B(p(u1 )) = {x1 , y1 }, we conclude x′2 = e = z1 . However, in the coloring φ we have φ(x′2 ) = φ(x1 ) 6= φ(z1 ) since α11 (a1 , b1 ) = α11 (a′2 , b′2 ), contradicting x′2 = z1 . This concludes the proof.  Claim 45. Let Σ ∈ Σ(ν14 ) and suppose that f1 , f2 , g1 , g2 are arcs of K̂Σ satisfying • u+ (f1 ) = u− (f2 ) • u+ (g1 ) = u− (g2 ) • q(f2 ) = q(g2 ). Then it also holds that q(f1 ) = q(g1 ). Proof. Recall that u− (f ) < q(f ) < u+ (f ) for every arc f of K̂Σ (by Claims 36 and 37). It follows from the assumptions that q(g1 ) < u+ (g1 ) = u− (g2 ) < q(g2 ) (35) and q(f1 ) < u+ (f1 ) = u− (f2 ) < q(f2 ) (36) in T . Thus q(g1 ) and q(f1 ) are comparable in T . Arguing by contradiction, suppose that q(f1 ) 6= q(g1 ). Using symmetry, we may assume without loss of generality q(f1 ) < q(g1 ) in T . Since q(g1 ) < q(g2 ) = q(f2 ) by (35) and u+ (f1 ) = u− (f2 ) < q(f2 ) in T by (36), the two nodes q(g1 ) and u+ (f1 ) are also comparable in T . First suppose that q(g1 ) < u+ (f1 ) in T . Then observe that the two nodes u+ (g1 ) and u+ (f1 ) are comparable in T since u+ (g1 ) = u− (g2 ) < q(g2 ) = q(f2 ) and u+ (f1 ) < q(f2 ) in T (by (35) and (36)). 34 Hence we have q(f1 ) < q(g1 ) < u+ (f1 ) 6 u+ (g1 ) or q(f1 ) < q(g1 ) < u+ (g1 ) 6 u+ (f1 ) in T , neither of which is possible by Claim 43, a contradiction. Next, assume that q(g1 ) > u+ (f1 ) in T . We immediately obtain u− (f2 ) = u+ (f1 ) 6 q(g1 ) < u+ (g1 ) < q(f2 ) in T , which is forbidden by Claim 44, again a contradiction.  Claim 46. The graph K̂Σ is bipartite for each Σ ∈ Σ(ν14 ). Proof. Suppose that there is an odd cycle C = {(ai , bi )}ki=1 in the undirected graph underlying K̂Σ . (Thus C is not necessarily a directed cycle.) For each i ∈ {1, . . . , k}, let fi be an arc between (ai , bi ) and (ai+1 , bi+1 ) in K̂Σ , where indices are taken cyclically as always. If fi = ((ai , bi ), (ai+1 , bi+1 )), that is, (ai , bi ) is the source of fi , we say that fi goes forward, while if fi = ((ai+1 , bi+1 ), (ai , bi )) we say that fi goes backward. We define a cyclically ordered sequence S of arcs in {f1 , f2 , . . . , fk } as follows. We start with S = (f1 , . . . , fk ). It will be convenient to say that we go along S in clockwise order whenver we use the forward direction in S (to not mix it up with forward and backward edges). Now, we repeat the following modification until it is no longer possible: If S has size at least 5 and there are two (cyclically) consecutive arcs f, f ′ in clockwise order in S with f going forward and f ′ going backward then remove both f and f ′ from S. By construction, the resulting sequence S has the following property: Either S contains at least five arcs and all arcs go in the same direction, or S contains exactly three arcs. We claim that during the above iterative process the cyclic sequence S fulfills the following invariants at all times: For any two consecutive arcs f and f ′ in clockwise order in S, (i) if f and f ′ both go forward then q(f ) < q(f ′ ) in T , while if f and f ′ both go backward then q(f ) > q(f ′ ) in T ; (ii) if f and f ′ go in the same direction then there exist arcs g, g ′ in K̂Σ such that • q(g) = q(f ); • q(g ′ ) = q(f ′ ), and • u+ (g) = u− (g ′ ) if f and f ′ go forward, u− (g) = u+ (g ′ ) otherwise, and (iii) if f and f ′ go in opposite directions then q(f ) = q(f ′ ). Note that (ii) implies (i). Indeed, suppose f and f ′ go forward (for the backward direction the argument is analogous). Then take arcs g and g ′ witnessing (ii). We have q(f ) = q(g) < u+ (g) = u− (g ′ ) < q(g ′ ) = q(f ′ ). (Recall that u− (g) < q(g) < u+ (g) for every arc g of K̂Σ by Claims 36 and 37.) First, we prove that the invariants hold at the beginning of the process, so for the sequence (f1 , . . . , fk ). In order to prove (ii), for each i ∈ {1, . . . , k} take g := fi and g ′ := fi+1 . Then clearly (ii) holds, and property (iii) follows from Claims 41 and 42. Next we show that the invariants hold after each modification step. Consider thus the sequence S just before a modification step, and suppose that S satisfied the required properties. Let f 0 , f 1 , f 2 , f 3 be the four consecutive arcs in S in clockwise order which are such that f 1 goes forward and f 2 goes backward. After removing f 1 and f 2 , the arcs f 0 and f 3 will become consecutive in S (in clockwise order). We only need to establish the invariants for the consecutive pair f 0 , f 3 , since all other consecutive pairs already satisfy them by assumption. Let us start with the case that f 0 and f 3 both go forward. Since f 1 , f 2 and f 3 alternate in directions, we get q(f 1 ) = q(f 2 ) = q(f 3 ) by (iii). By (ii) and the fact that f 0 and f 1 go forward, there are arcs g 0 , g 1 in K̂Σ such that q(g 0 ) = q(f 0 ), q(g 1 ) = q(f 1 ) and u+ (g 0 ) = u− (g 1 ). Now, since q(g 1 ) = q(f 1 ) = q(f 3 ), the arcs g 0 , g 1 also fulfill the conditions of (ii) for f 0 and f 3 . The case that both f 0 and f 3 go backward is symmetric to the previous one and is thus omitted. Next, suppose that f 0 goes forward and f 3 goes backward. Here we have to show that (iii) holds for 0 f and f 3 . Since f 0 and f 1 both go forward and f 2 and f 3 both go backward, by (ii) there are arcs g 0 , g 1 and g 2 , g 3 in K̂Σ such that q(g j ) = q(f j ) for each j ∈ {0, 1, 2, 3}, u+ (g 0 ) = u− (g 1 ), and u− (g 2 ) = u+ (g 3 ). Using (iii) we deduce that q(g 1 ) = q(f 1 ) = q(f 2 ) = q(g 2 ). Applying Claim 45 on the arcs g 0 , g 1 , g 3 , g 2 (in this order), we conclude q(f 0 ) = q(g 0 ) = q(g 3 ) = q(f 3 ), as desired. Finally, assume that f 0 goes backward and f 3 goes forward. Again we have show that (iii) holds for 0 f , f 3 . But in this case the four directions of f 0 , f 1 , f 2 , f 3 alternate. It follows that q(f 0 ) = q(f 1 ) = q(f 2 ) = q(f 3 ) by (iii). Now that the above invariants of S have been established, let us go back to the final sequence S resulting from the modification process. We claim that there are always two consecutive arcs going in opposite directions in S. Indeed, if not then they either all go forward or all go backward. In the first 35 case q(f ) < q(f ′ ) in T for any two consecutive arcs f, f ′ in clockwise order in S by (i), while in the second case q(f ) > q(f ′ ) in T for any two such arcs f, f ′ . However, neither of these two situations can occur in a circular sequence. This shows in particular that the modification process results in a sequence S of size 3, say S = (f 1 , f 2 , f 3 ). We may suppose without loss of generality that f 1 and f 2 go in the same direction and f 3 in the other (since the sequence S can always be shifted cyclically to ensure this property). This implies q(f 1 ) 6= q(f 2 ) by (i), q(f 2 ) = q(f 3 ) by (iii), and q(f 3 ) = q(f 1 ) by (iii). This is a contradiction, which concludes the proof.  Using Claim 46 we let ψ14,Σ : MM(P, ν14 , Σ) → {1, 2} be a 2-coloring of K̂Σ , for each Σ ∈ Σ(ν14 ). The function α14 then records the color of a pair in this coloring: For each Σ ∈ Σ(ν14 ) and each pair (a, b) ∈ MM(P, ν14 , Σ), we let α14 (a, b) := ψ14,Σ (a, b). 3.13. Fourth leaf of Ψ: Node ν15 . It remains to verify that for each Σ ∈ Σ(ν15 ) the set MM(P, ν15 , Σ) is reversible. Recall that α13 ensures that there are no 2-cycles in MM(P, ν15 , Σ) and that α14 ensures that there are no strict alternating cycles of length at least 3 in MM(P, ν15 , Σ). It follows that MM(P, ν15 , Σ) is reversible. This concludes the proof of Theorem 7. Acknowledgments This research was initiated during the workshop Order and Geometry held at the Technische Universität Berlin in August 2013. We are grateful to the organizers and the other participants for providing a very stimulative research environment. We also thank Grzegorz Gutowski and Tomasz Krawczyk for many fruitful discussions at the early stage of this project. Finally, we are much grateful to the referees for their many helpful remarks and suggestions, which greatly improved the readability of the paper. References [1] Csaba Biró, Mitchel T. Keller, and Stephen J. Young. Posets with cover graph of pathwidth two have bounded dimension. Order, in press, DOI 10.1007/s11083-015-9359-7. http://arxiv.org/abs/1308.4877. [2] David Eppstein. Diameter and treewidth in minor-closed graph families. Algorithmica, 27(3-4):275–291, 2000. [3] Stefan Felsner, William T. Trotter, and Veit Wiechert. The dimension of posets with planar cover graphs. Graphs Combin., 31(4):927–939, 2015. [4] Gwenaël Joret, Piotr Micek, Kevin G. Milans, William T. Trotter, Bartosz Walczak, and Ruidong Wang. Tree-width and dimension. Combinatorica, in press, DOI 10.1007/s00493-014-3081-8. http://arxiv.org/abs/1301.5271. [5] Gwenaël Joret, Piotr Micek, and Veit Wiechert. Sparsity and dimension. In Proceedings of the Twenty-Seventh Annual ACM-SIAM Symposium on Discrete Algorithms, pages 1804–1813, 2016. http://arxiv.org/abs/1507.01120. [6] David Kelly. On the dimension of partially ordered sets. Discrete Math., 35:135–156, 1981. [7] Piotr Micek and Veit Wiechert. Topological minors of cover graphs and dimension. Submitted, http://arxiv.org/ abs/1504.07388. [8] Noah Streib and William T. Trotter. Dimension and height for posets with planar cover graphs. European J. Combin., 35:474–489, 2014. [9] William T. Trotter. Combinatorics and partially ordered sets. Dimension theory. Johns Hopkins Series in the Mathematical Sciences. Johns Hopkins University Press, Baltimore, MD, 1992. [10] William T. Trotter. Partially ordered sets. In Handbook of combinatorics, Vol. 1, 2, pages 433–480. Elsevier Sci. B. V., Amsterdam, 1995. [11] William T. Trotter, Jr. and John I. Moore, Jr. The dimension of planar posets. J. Combinatorial Theory Ser. B, 22(1):54–67, 1977. [12] Bartosz Walczak. Minors and dimension. In Proceedings of the Twenty-Sixth Annual ACM-SIAM Symposium on Discrete Algorithms, pages 1698–1707, 2015. http://arxiv.org/abs/1407.4066. 36 (G. Joret) Computer Science Department, Université Libre de Bruxelles, Brussels, Belgium E-mail address: [email protected] (P. Micek) Theoretical Computer Science Department, Faculty of Mathematics and Computer Science, Jagiellonian University, Kraków, Poland E-mail address: [email protected] (W. T. Trotter, R. Wang) School of Mathematics, Georgia Institute of Technology, Atlanta, Georgia 30332, U.S.A. E-mail address: [email protected], [email protected] (V. Wiechert) Institut für Mathematik, Technische Universität Berlin, Berlin, Germany E-mail address: [email protected] 37