Sublinear Extensions of Polygons

SUBLINEAR EXTENSIONS OF POLYGONS

YAROSLAV SHITOV

Abstract. Every convex polygon with n vertices is a linear projection of a higher-dimensional polytope with at most 147 n2/3 facets.

We revisit the following appealing question, which survived a thorough discussion in recent literature without receiving any deﬁnitive progress towards its resolution. Question 1. What is the smallest number, pc(n), such that every convex polygon with n vertices is a linear projection of some polytope with at most pc(n) facets? Apart from being a natural question to ask in polyhedral combinatorics, Ques- tion 1 is being studied from the point of view of modern optimization theory and linear algebra. Our approach is purely geometric, but still we need to recall several concepts of optimization theory to explain the relevance of Question 1.

1. Introduction Let P be a polytope, that is, the convex hull of a ﬁnite collection of points in a d-dimensional Euclidean space. An extended formulation of P is a pair (Q, π) consisting of a polytope Q ⊂ Rm and a linear projection π : Rm → Rd such that π(Q) = P . The number of facets of such a polytope Q is called the size of the formulation (Q, π), and the extension complexity of P is the smallest possible size of any extended formulation of P . This quantity, further denoted by xc(P ), equals the smallest possible number of inequalities in a representation of any polytope that can be sent to P by a linear projection [17, 60]. As we see, Question 1 is devoted to the worst-case extension complexities of two-dimensional polytopes, so we decided to use the notation ‘pc’ as an abbreviation of ‘polygon complexity.’ Extended formulations became a prominent technique in optimization after a 1991 paper by Yannakakis [60], who discussed a possibility of building fast al- gorithms for hard combinatorial problems by constructing smaller extended formulations to linear programs corresponding to these problems. Several intrigu- ing questions of this nature were solved quite recently, which include the lack of arXiv:1412.0728v2 [math.CO] 29 Feb 2020 polynomial-size extended formulations for the polytope associated to the traveling salesman problem [18]. Another notable result [49] gives an exponential lower bound for the extension complexity of the matching polytope, which is in contrast to the existence of a polynomial-time algorithm solving the matching problem [20]. Other results on the topic include strong lower bounds for the complexities of the polytopes associated to diﬀerent exact and approximate optimization problems such as max-cut [9], independent set [24], knapsack [46] and many others [2]. The extension complexity is also being discussed for polytopes not necessarily arising from particular combinatorial problems, including those with few vertices and facets [28, 29, 43],

2000 Mathematics Subject Classification. 52B05, 52B12, 15A23. Key words and phrases. Convex polytope, extended formulation, nonnegative matrices. 1 2 YAROSLAV SHITOV the correlation polytope [35], the permutahedron [25], 0-1 polytopes [8, 48], two- level polytopes [1], orbitopes [15], Cartesian products and hypersimplices [31] and others. These also include a prominent example of polygons [5, 19, 50, 55], which this paper is devoted to and which have ‘prototypical importance’ to applications, according to Braun and Pokutta [6]. Also, the study of extended formulations has been developed to reflect the case of semidefinite programming [16, 27, 30], and it may require tools of linear algebra [60], information theory [7], communication complexity [14], quantum learning [38], tropical mathematics [51] and other fields; an interested reader is referred to survey papers [12, 34] for further details. Yannakakis [60] developed a linear algebraic approach to extended formulations based on nonnegative matrices, that is, matrices with nonnegative real entries. The nonnegative rank of such a matrix M is the smallest integer k for which M can be written as a sum of k nonnegative rank-one matrices. Now consider a polytope P , with v vertices and f facets, defined as a subset of Rd by the conditions fε(x) = αε, gϕ(x) > βϕ, where the letter ε indexes a finite set of linear equations, the ϕ runs over the facets d of P , and the mappings fε and gϕ are linear functionals R → R.A slack matrix of P can be defined by the formula

Sνϕ = gϕ(ν) − βϕ with rows and columns of S labeled by the vertices and facets of P , respectively. It is not hard to see that the rank of S equals dim P +1; Yannakakis [60] proves that the nonnegative rank of S equals the extension complexity of P . We note in passing that the nonnegative rank is an important concept of linear algebra in its own right [21, 57] and for theoretical studies in demography, quantum mechanics [11], and statistics [37, 42]. Also, this notion is relevant for several real-life challenges like signal processing [23], text mining [10], and image processing [39]. Theorem 2. (See [60].) The extension complexity of a polytope P equals the nonnegative rank of any slack matrix of P . Corollary 3. For n > 3, the value pc(n) equals the largest nonnegative rank of a nonnegative matrix with at most n rows and conventional rank three. Proof. This follows from Theorem 2 by standard techniques. In particular, we can realize pc(n) as the nonnegative rank of an n × n matrix of rank three by a direct application of this result to a slack matrix of a convex n-gon with extension complexity pc(n). The fact that the nonnegative rank of any n×m nonnegative matrix A with rank three cannot exceed pc(n) also follows from Theorem 2, as outlined in the proof of Theorem 3.1 in [50]. To give a sketch of the proof, we consider such a matrix A, and we denote by ∆ ⊂ Rn the simplex consisting of nonnegative vectors whose coordinates sum to 1. Since ∆ has n facets, the intersection of ∆ with the column space of A is a polygon P with k 6 n vertices. Denoting by S the matrix of the column coordinate vectors of the vertices of P , we have A = SB with B nonnegative. Since S is a slack matrix of P , it has nonnegative rank at most pc(n), and so does A. 2. Polygons on a plane Question 1 takes an important place in the study of extended formulations. A detailed account of this problem was carried out by the authors of [19], who adopted SUBLINEAR EXTENSIONS OF POLYGONS 3 √ the dimension counting method and gave an Ω( n) lower bound for the extension complexity of a generic convex n-gon. More presicely, they obtained the inequalities √ (2.1) 2n 6 pc(n) 6 n, in which the upper bound is trivial. Question 1 and its analogues were studied in at least a dozen of papers [3, 22, 27, 30, 33, 40, 43, 44, 45, 50, 51, 53, 54, 56] and mentioned in several dozens of other works, including a number of textbooks, surveys and highly cited research papers [7, 12, 17, 41, 47, 48, 58]. This question was discussed at open problem sessions of computer science conferences [4, 36] and online in popular media [13, 26], but it has seen√ no progress on either bound except for improving the constant factors in front of n and n. Concerning the asymptotic behavior of pc(n), the authors of the initial paper [19] and most other experts seemed to expect that the upper bound in (2.1) is closer to the actual value. Braun and Pokutta [7] point out the importance of the following question and mention it alongside a list of major problems in the theory of extended formulations.

Question 4. Do we have pc(n) > εn for some fixed ε > 0? This question appeared in online media such as the Open Problem Garden [13] and MathOverflow [26]. An affirmative answer to Question 4 was conjectured in [3] and claimed to be proven in [40]. However, Hrubeˇs[33] showed that the argument of [40] is flawed and formulated a problem equivalent to Question 4; a further version of this question but specified to ε = 0.5 appears in [56]. The authors of [19] expected an affirmative answer to another version obtained by replacing εn with Ω(˜ n) in the formulation of Question 4, which would mean that the upper bound in (2.1) is tight up to logarithmic factors. Contrary to these expectations, we show that the lower bound in (2.1) is closer to the actual value of pc(n) in the asymptotical sense.

2/3 Theorem 5. We have pc(n) 6 147 n for all n > 3. We close the preliminary part of our paper with a more detailed description of the progress achieved on Questions 1 and 4 before the work we present. As far as we know, Beasley and Laffey [3] were the first to consider this question; they followed the linear algebraic approach in terms of Corollary 3. They proved that a special family of nonnegative n×n rank-three matrices, called Euclidean distance matrices, have nonnegative rank at least log n, and conjectured that the maximal value of this rank is n. In our notation, their result stated that pc(n) > log n, and the conjecture was pc(n) = n. A subsequent paper [40] claimed to prove this conjecture, but later works [22, 33] pointed out a flaw in their proof. The authors of [22] saved a part of the argument in [40] and showed that the so-called restricted nonnegative rank of Euclidean distance matrices equals n. They reiterated the question as to whether the equality pc(n) = n holds for general n and proved it for n not exceeding five. The n = 6 case of this question was treated in the affirmative in [27], but already for n = 7 the answer turned out to be negative — the equality pc(7) = 6 was proved in [44, 50]. As a corollary of the latter result, the authors of [44, 50] get the estimate pc(n) 6 d6n/7e for all n, which stood as the best known upper bound on pc(n) to this date. A more thorough description of the n = 7 case lead the authors of [45] to a family of polytopes whose optimal extended formulations require the use of hidden vertices. The tropical approach to Questions 1 and 4 does not look promising; the author of [53, 54] gave new combinatorial proofs that pc(6) = 6 and pc(8) = 7, but both of these equalities were known at the time because an improved 4 YAROSLAV SHITOV √ counting argument of Padrol [43] confirmed the inequality pc(n) > 2 2n − 2 − 1 for all n. An algorithmic treatment of these questions gave the authors of [56] some important information on polygons whose vertices are randomly chosen on a circle, and they went on to conjecture that pc(n) 6 0.5n + 3 for all n. The paper [30] showed that any convex n-gon admits a semidefinite extended formulation of size at most 4dn/6e, but this did not imply any progress on our pc function because the semidefinite approach has more expressive power in comparison to the extended formulations as in the setting of this paper. Nevertheless, Question 4 remained open even in the case of stronger, semidefinite, extended formulations.

3. Preliminaries This paper is devoted to the proof of Theorem 5. In this section, we collect several standard notational conventions and basic results used in the course of our discussion, and we give a short overview of our approach. We work in the Euclidean real space Rd with the conventional Euclidean metric. The notation dist(u, v) stands for the distance between two points u, v ∈ Rd, and the distance dist(U, V ) between two non-empty subsets U, V ⊂ Rd is the infimum of all possible distances dist(u, v) with u ∈ U and v ∈ V . The interior of a set V ⊂ Rd is understood in terms of the topology inherited from the Euclidean metric, and the relative interior of a segment σ ⊂ Rd is the same segment but with endpoints removed. A set V ⊂ Rd is called convex if for any pair of distinct points u, v ∈ V , the segment between u and v is contained in V . The convex hull of a set S ⊂ Rd is denoted by conv S, and it is defined as the intersection of all convex sets in Rd containing S. We define a polytope in Rd and a polygon in R2 as the convex hull of a finite collection of points, that is, we do not work with non-convex generalizations −→ of polytopes here. If u, v ∈ Rd are distinct points, then we write uv for the oriented segment with the beginning at u and ending at v. We note that this is different from saying that uv−→ is a vector, which would rather mean the class of all segments with the same length and direction. A significant part of our considerations goes over the two-dimensional space, so we give several definitions that we mostly use in this setting. A ray ρ ⊂ R2 is a closed convex unbounded subset of a straight line; if a is the apex and b is a non- apex point on ρ, then we say that ρ goes from a towards b. If r, s are two oriented segments or rays, then we write ∠(r, s) for the measure of the angle between them, −→ −→ and we also write ∠uvw for ∠(vu, vw) whenever u, v, w are points.

Observation 6. If a1, a2, a3 are oriented segments or rays, then

∠(a1, a2) + ∠(a2, a3) > ∠(a1, a3). 2 If P ⊂ R is a polygon, then the turning angle at a vertex v is π − ∠v−vv+, where v− and v+ are two vertices adjacent to v in P . The turning angle of an edge e of a polygon is the sum of the turning angles at the two endpoints of e. Deﬁnition 7. We write u ∧ v to denote (1) the straight line connecting u, v if they are distinct points, (2) the intersection of u, v if they are non-collinear straight lines on a plane. We proceed with two basic results on the extension complexity. Lemma 8. Let P and Q be polytopes in Rd each of which is diﬀerent from a single point. Then xc(conv P ∪ Q) 6 xc(P ) + xc(Q). SUBLINEAR EXTENSIONS OF POLYGONS 5

Proof. See Proposition 3.1.1 in [59] and a similar Proposition 2.8(6) in [27].

Observation 9. Let P ⊂ Rd be a polytope, and let H ⊂ Rd be a closed half-space. If the intersection P ∩ H is non-empty, then xc(P ∩ H) 6 xc(P ) + 1. Proof. Let Q ⊂ Rd ×Rα be a polytope with xc(P ) facets such that π(Q) = P , where π is the projection onto the first d coordinates. Then the polytope Q0 defined as α 0 Q ∩ (H × R ) has at most xc(P ) + 1 facets, and we have π(Q ) = P ∩ H. The general strategy of our approach is as follows. The forthcoming Sections 4– 6 reduce the question of constructing a small extended formulation of a polygon to a specific problem in plane geometry, which comes from a certain special class of three-dimensional extensions. This class is introduced in Section 4 under the name of acute polyhedra, and their Schlegel diagrams are investigated in Section 5. Section 6 contains Theorem 28, which is the main result of the first part of the paper, and it explains how to glue several extensions coming from acute polytopes together to get a higher dimensional extended formulation of an initial polygon. In view of Lemma 8, the progress on Question 1 can be made by extracting a sufficiently large subset S of the vertices of a given polygon for which the extension complexity of conv S is small, so the subsequent Sections 7–11 are devoted to extracting such a subset which would be well behaved with respect to the methods developed in Sections 4–6. More precisely, Sections 7 and 8 give further notational conventions needed to get an appropriate two-dimensional description of polygons that allow the use of Theorem 28. In Section 9, we introduce a further auxiliary notion of when the sequence of the vertices of a given polygon is slanted, and we explain that a polygon without large slanted subsequences should have relatively small extension complexity. The problem reduces to proving good upper bounds of polygons with slanted vertex sequences, which is done in the technical Sections 10 and 11. Our proof is completed in Section 12, and we give several additional remarks and suggestions on further research on this topic in Section 13.

4. Acute polyhedra In this section, we introduce the class of acute polytopes, which are useful for constructing small extended formulations. We recall that, for three-dimensional polyhedra, the angle between two faces A and B with a common edge e equals the angle between two oriented segments that lie on A and B, respectively, have their beginnings on e, and are orthogonal to e.

Definition 10. Assume P ⊂ Rd is a polytope and B is one of its facets. If any other facet F 6= B of P satisfies the conditions that (a) dim B ∩ F = dim P − 2 and (b) the angle between B and F is acute, then P is called an acute polytope with base facet B. Remark 11. Our notation is suggested by the fact that an acute triangle satisfies the above conditions whichever side of it we choose as the base. Right and obtuse triangles can also become ’acute polytopes’ in our sense, but the only possible bases are their longest sides. For n > 3, no n-gon on the plane can satisfy Definition 10 because of the condition (a). In our further considerations, we restrict our attention to three-dimensional acute polytopes, and we call them acute polyhedra. 6 YAROSLAV SHITOV

One example of an acute polyhedron is given on Figure 1 below.

Observation 12. Let P ⊂ R3 be an acute polyhedron with base facet B. Assume that B lies on the plane {z = 0} and its edge-deﬁning inequalities are {aix + biy + ci > 0} with i = 1, . . . , k. Then the non-base facets of P are deﬁned by

aix + biy + ci > εiz, where ε1, . . . , εk are non-zero numbers with the same sign. Proof. According to the part (a) of Definition 10, every non-base facet of P passes through some edge of B. Any such facet should have the inequality as in the displayed formula, and the part (b) of Definition 10 shows that the ε’s should be taken positive if P ⊂ {z > 0} and negative if P ⊂ {z 6 0}. We need one more concept to be used in further considerations. Definition 13. We say that an edge e of an acute polyhedron with base B is main if exactly one of the endpoints of e lies on B. Lemma 14. Any base vertex of an acute polyhedron belongs to a unique main edge. Proof. It is clear from Definition 13 that a base vertex v is adjacent to two other base vertices, so if the current statement was false, there would be at least four vertices adjacent to v. This means that there would be at least four faces containing v, in which case we can find three such faces ϕ1, ϕ2, ϕ3 none of which is the base. The part (a) of Definition 10 shows that these faces pass through base edges e1, e2, e3, respectively. If any such ei did not contain v, the face ϕi would pass through three non-collinear base vertices, which is a contradiction because ϕi is not the base. Therefore, the base edges e1, e2, e3 pass through v, which means that two of them coincide. This implies that the base and two of the faces ϕ1, ϕ2, ϕ3 have a common edge, which is impossible for three-dimensional polyhedra. 2 Lemma 15. Let V ⊂ R be a convex polygon with vertices v0, v1, . . . , vn, and let y1, . . . , yn be a family of inner points of V . Then there is an acute polyhedron P ⊂ R3 with base V such that, for any i ∈ {1, . . . , n}, the image of the main edge passing from vi under the orthogonal projection of P onto V is collinear to vi ∧ yi.

Proof. We embed the plane containing V into R3 and choose one of the two resulting half-spaces U to be called upper. For all i ∈ {1, . . . , n}, we define Hi as the plane that passes through vi, yi and is orthogonal to the plane of V . Also, we denote the edges of V consecutively as e0, . . . , en, and we assume that ej has vj and vj+1 as endpoints, where vn+1 stands for v0 and j ranges in 0, . . . , n. In the following construction of an acute polyhedron P ⊂ U, the notation Fj stands for the upper half-plane that contains ej and the corresponding non-base facet of P . We choose F0 as an arbitrary upper half-plane passing through e0 and having an acute angle with V , and, by the formulation of the lemma, we conclude that the main edge passing through v1 should be a part of the ray F0 ∩ H1, which allows us to define F1 as the upper half-plane containing e1 and F0 ∩ H1. We proceed by the induction, and we define Fi as the upper half-plane passing through ei and Fi−1 ∩ Hi for all i ∈ {1, . . . , n}. It remains to note that Fi forms an acute angle with V because it contains a non-apex point on Fi−1 ∩ Hi which appears inside V under the orthogonal projection onto the plane containing V . SUBLINEAR EXTENSIONS OF POLYGONS 7

5. Acute diagrams Our further considerations need a more accurate characterization of acute polyhedra in terms of their Schlegel diagrams [32]. In this section, we describe the class of planar straight-line graphs that can arise as the diagrams of acute polyhedra. Observation 16. Let P ⊂ R3 be an acute polyhedron; consider the orthogonal projection π onto the plane containing its base B. Then (a) the non-base points of P project into the interior of the base, and (b) the mapping π is injective on the non-base boundary points of P . Proof. The conclusion (a) is immediate from the part (b) of Definition 10. In order to prove (b), we consider two non-base points u 6= v on the boundary of P with π(u) = π(v). Since B is a facet, one of the points u, v should lie between the plane of B and the other point. If this middle point is u, then by the item (a) it lies in the interior of conv B ∪{v} and hence in the interior of P , which is a contradiction. Remark 17. In other words, the projection π as in Observation 16 gives a Schlegel diagram of P . This diagram is to be called the acute diagram of P relative to the base facet B, or just the acute diagram of P if the choice of the base is clear from context. We also say that P is an acute lifting of the corresponding acute diagram. Figure 1 gives an example of an acute polytope and the corresponding diagram. Definition 18. A planar straight-line graph ∆ is a topological graph in R2 whose arcs are non-crossing line segments, that is, any pair of arcs are either disjoint or have a common node. The convex hull of ∆ is to be called the base of ∆, and the vertices of the base are called the base nodes of ∆. We say that line segments a, b, c ⊂ R2 are concurrent if the straight lines containing a, b, c are concurrent, that is, are either parallel or have a common point. Lemma 19. Let ∆ be the acute diagram of an acute polyhedron P with base B. Then ∆ is a planar straight-line graph such that (o) the base of ∆ is B, (i) every node of ∆ has degree at least three, (ii) the non-base nodes of ∆ lie in the interior of the base, (iii) every edge of the base is an arc of ∆, (iv) every bounded face f of ∆ contains exactly one arc ef of the base, (v) if a non-base arc e of ∆ separates faces f, g, then e, ef , eg are concurrent. Proof. The planarity of ∆ and the item (i) follow from the part (b) of Observa- tion 16, and the items (o, ii, iii) can be obtained from the part (a) of Observation 16. To prove the item (iv), we note that any face of P contains at least one edge of the base by the part (a) of Definition 10, and a face with at least two such edges should be the base itself. Now we proceed with the item (v). Let ϕ, ψ be two faces of P with a common edge ε. We denote by B the base plane of P , and by Φ, Ψ the two planes containing ϕ, ψ, respectively. Then Φ ∩ Ψ is the straight line containing ε, and its intersection with the base plane is Φ ∩ Ψ ∩ B, which is the same as the intersection of the two straight lines Φ ∩ B,Ψ ∩ B containing the base edges of ϕ, ψ. Remark 20. Let ∆ be a planar straight-line graph satisfying the conditions (i)–(v) as in Lemma 19, and let s be an arbitrary base node of ∆. We define ∆in as the 8 YAROSLAV SHITOV graph obtained by removing all the base arcs from ∆. According to the item (iv) of Lemma 19, this graph is a tree, so we can define the s-orientation by saying that an arc conv{u, v} of ∆in is oriented from u to v if and only if ρ(u, s) > ρ(v, s), where ρ is the graph-theoretical geodesic distance function on ∆in.

Figure 1. An acute polytope, its acute diagram, and an s-orientation.

Now we need to explain how to lift an acute diagram into an acute polytope. Observation 21. In the s-orientation as in Remark 20, every node diﬀerent from s has outdegree 1. The node s is a sink, that is, it has outdegree 0. Proof. Since ∆in is a tree, the shortest path between any pair of nodes is unique, which proves the ﬁrst statement. The conclusion on the outdegree of s is trivial. Lemma 22. Let u be an inner node of the s-orientation of ∆ as in Remark 20. Let Λ(u) be the set of all base nodes from which u can be reached by oriented paths. Then Λ(u) is a set of consecutive nodes of the base.

Proof. For v1, v2 in Λ(u), we can construct oriented paths π1, π2 beginning at v1, v2, respectively, and ending at u. The union of π1 and π2 splits ∆ into two connected components, and for any base node x outside the component of the sink, any oriented path from x to s passes through u. Lemma 23. Any planar straight-line graph ∆ satisfying the conditions (i)–(v) as in Lemma 19 contains a triangle formed by two main arcs and one base arc e. This arc e can be chosen so that it does not contain a base node s fixed in advance. Proof. We choose a node x furthest from s in terms of the graph-theoretic geodesic distance on ∆in. Since ∆in is a tree, this x should be its leaf, that is, a base node with a unique main arc α leading from x to some inner node y. By the condition (i) in Lemma 19 and Observation 21, we see that y is adjacent to at least two nodes x, x0 at the maximal distance from s. According to Lemma 22, the base nodes from which y can be reached are consecutive, and any such node should be connected with y by an arc because of the distance maximality assumption. Lemma 24. Let ∆ be a planar straight-line graph satisfying the conditions (i)–(v) as in Lemma 19. If the base of ∆ is not a triangle, then there is a base arc e with turning angle at most π which forms a triangle together with two main arcs adjacent to the endpoints of e. If, additionally, the base of ∆ is not a trapezoid, then this arc e can be chosen to have the turning angle strictly less than π. Proof. Let us call bad those arcs of the base of ∆ that have turning angle π or greater. If all bad arcs are adjacent to a common node s, then the proof is completed SUBLINEAR EXTENSIONS OF POLYGONS 9 by the application of Lemma 23. If there are two non-adjacent bad arcs, then the total of the turning angles at all four of their endpoints is 2π, so the base of ∆ has no other nodes except these four, so its further analysis is straightforward. Lemma 25. A planar straight-line graph ∆ satisfying the conditions (i)–(v) as in Lemma 19 can be obtained as an acute diagram of some acute polyhedron P . Proof. We work by the induction on k, the number of nodes of the base of ∆. The base of the induction comes from the two cases in which Lemma 24 is not applicable, namely, the triangle and trapezoid bases. These cases are easy because the triangle bases come from triangular pyramids, and the trapezoid case comes from cutting the triangular prism from the two sides as in Figure 1. We proceed with an application of Lemma 24, and we find a base arc e = conv{x1, x2} with turning angle less than π such that both x1, x2 are adjacent to the same inner node y. Let ε1, ε2 be the base arcs that are different from e and 0 contain x1, x2, respectively. Now to construct a new graph ∆ from ∆, we (1) continue ε1, ε2 to the intersection point x, (2) connect x to y, and erase the arcs conv{x1, y} and conv{x2, y}, 0 and we do not call x1 and x2 the nodes of ∆ . If the degree of y became less than three after the transformations (1)–(2), then we do not call it a node as well.

Figure 2. The inductive step in the proof of Lemma 25.

It is straightforward to see that the graph ∆0 is planar and straight-line, and it satisfies the conditions (i)–(v) as in Lemma 19. Using the inductive assumption, we build an acute lifting P 0 of ∆0, and it remains to cut off a piece of P 0 by the plane 0 passing through x1, x2, y0, where y0 is the point on the main edge of P passing 0 from x that maps to y under the orthogonal projection of P onto its base. We finalize the section with a more delicate general property of acute diagrams.

Corollary 26. Let V be a convex polygon with vertices v0, v1, . . . , vn, and let y1, . . . , yn be a family of inner points of V . Then there is an acute diagram with base V such that, for any i ∈ {1, . . . , n}, the main edge from vi lies on vi ∧ yi. Proof. This a reformulation of Lemma 15. 6. Glueing acute extensions together Now we are going to employ the approach of acute polyhedra in a construction of small extended formulations. We recall that a polyhedral cone in Rd is the convex hull of a ﬁnite collection of rays passing from the same apex O. Such a cone is called pointed if it does not contain a straight line. 10 YAROSLAV SHITOV

Observation 27. Let C ⊂ Rq be a pointed cone deﬁned as the convex hull of its q extreme rays l1, . . . , lk passing from the apex O. Let H ⊂ R be a closed half- space containing an unbounded part of every l1, . . . , lk such that the boundary of H q d intersects li at a unique point vi for all i ∈ {1, . . . , k}. Also, let π : R → R be a linear projection. Then π(H ∩ C) is the convex hull of the union of the rays passing from π(vi) towards the direction of π(li) taken over all i = 1, . . . , k.

Proof. The ray H ∩ li passes from vi in the direction of li, and the set H ∩ C is the convex hull of the union of all H ∩ li over i = 1, . . . , k. Since the mappings π and q conv on the subsets of R commute, we get the desired result. Now we are ready to explain how to glue several three-dimensional acute extensions together to get quite a small higher-dimensional extended formulation.

Figure 3. An application of Theorem 28: Two acute diagrams conﬁrm that the 14-gon on the right has xc at most 10.

Theorem 28. Let P be a convex polygon on a plane with a distinguished non- empty set S of the set V of all vertices. Assume that, for any s ∈ S, one chooses two vertices s0, s00 on two edges of P adjacent to s, and one picks a set {s1, . . . , sδ} 0 00 consisting of δ > 1 new points in the interior of the triangle Ts = conv{s, s , s }. 0 00 Assume that the choice of s , s was made so that the triangles Ts are disjoint for different s. Assume that, for any i ∈ {1, . . . , δ}, there exists an acute diagram Di with base face P such that, for any s ∈ S, the segment between s and si is a subset of the main edge of Di passing from s. Then the convex hull of ! [ (6.1) {s0, s00, s1, . . . , sδ} ∪ (V \ S) s∈S SUBLINEAR EXTENSIONS OF POLYGONS 11 is a polygon of extension complexity not exceeding |V | + |S| + δ. Proof. We are going to construct a polytope P0 with at most |V | + |S| + δ facets in R2 ⊕ Rδ = {(x, y, z1, . . . , zδ)} such that the projection of P0 onto the first two coordinates is the desired set, that is, the convex hull of (6.1). Let aex+bey+ce > 0 be a defining inequality corresponding to an edge e of the polygon P . According to Observation 12, the lifting of the diagram Di gives an acute polytope Ai ⊂ i i {(x, y, z )} defined by the conditions z > 0 and i i aex + bey + ce > εez with the ε’s being positive real numbers. We define P ⊂ R2 ⊕ Rδ by the conditions i (6.2) z > 0,

1 1 δ δ (6.3) aex + bey + ce > εez + ... + εez , where i ranges in {1, . . . , δ}, and e runs over all edges of P . It is clear that dim P = δ + 2, and we use the part (a) of Observation 16 to see that π(P) = P , where 2 d 2 π : R ⊕R → R is the projection onto {(x, y)}. Now we pick any s = (xs, ys) ∈ S, and we see that the point σ = (xs, ys, 0,..., 0) is the vertex of P, which fact is easy because σ is the vertex of the face of P defined by z1 = ... = zδ = 0. Since there are exactly δ + 2 facets of P intersecting at σ, namely all those in (6.2) and those in (6.3) corresponding to s ∈ e, we conclude that the vertex figure of σ is a simplex. In particular, the cone corresponding to σ is defined by δ + 2 extreme rays, two of which correspond to the two edges of P adjacent to s and the other δ are 1 i i (6.4) (xs + αst, ys + βst, 0,..., 0, t, 0,..., 0), t > 0, i i where we have i − 1 zeros in front of t, and (αs, βs) is a vector pointing to the i 2 δ direction from s towards s . Now we define the half-space Hs ⊂ R ⊕ R which (1) does not contain σ, (2) contains (xs0 , ys0 , 0,..., 0) and (xs00 , ys00 , 0,..., 0), (3) for any i = 1, . . . , δ, it contains the point on the ray (6.4) that projects to si. 0 We can define P as the intersection of P and all Hs with s ∈ S, and we conclude that π(P) is the desired polygon by applying Observation 27. 7. Thin sequences We proceed with several further notational conventions. A sequence v = (v1, . . . , vn) of distinct points on a plane is called correct if these points are the vertices of their convex hull P and the segment between any pair of consecutive points in v is an edge of a polygon P . A correct sequence is called a cw-sequence if the order of the vertices is clockwise, and it is called a ccw-sequence otherwise.

Deﬁnition 29. Let α ∈ (0, π) and n > 3. A sequence v = (v1, . . . , vn) is called α-thin if v is correct and, additionally, the turning angle of the edge conv{v1, vn} in the polygon conv v is greater than 2π − α, that is,

∠vnv1v2 + ∠v1vnvn−1 < α. We say that v is thin if it is α-thin for some α ∈ (0, π).

1We note that a ray of the form (6.4) contains an edge of P because it contains, by the deﬁnition of an acute diagram, an edge of the corresponding acute polyhedron Ai, and this polyhedron is a face of P deﬁned by the conjunction of the equations zj = 0 with j 6= i. 12 YAROSLAV SHITOV

In the following easy observation, we use the total turning angle as a parameter of the boundary of a given convex polygon P to split the vertices of P into the union of several thin sequences.

Figure 4. Splitting the vertices of a polygon into thin sequences.

Observation 30. Let P be a convex polygon on n vertices, and let q > 3 be an integer. Then the vertices of P can be partitioned into at most q sets each of which is either a point or a pair of points, or else a set that forms a 2π/q-thin sequence.

Proof. We consider a consecutive enumeration e1, . . . , en of the relative interiors of the edges of P . We have

0 = τ(e1) < τ(e2) < . . . < τ(en) < 2π, where τ(ei) is the sum of the turning angles at all the vertices that lie between e1 and ei in the sense of our enumeration. We also denote by en+1 a copy of e1, and we assume τ(en+1) = 2π. For j ∈ {1, . . . , q}, we deﬁne µj as the smallest index for which τ(eµj ) > 2πj/q; we also set µ0 = 1. The desired partition consists of the q sets some of which may be empty, and the jth of these sets consists of the vertices that lie between eµj−1 and eµj again with respect to our enumeration. Observation 30 shows that, for constant α, a worst-case upper bound on the extension complexity of an α-thin sequence gives an upper bound on pc(n) which is at most a constant multiple worse than the bound in the α-thin case. So we switch our attention to thin sequences to return to general polygons in Section 12.

Deﬁnition 31. Let v = (v1, . . . , vn) be a thin cw-sequence. In what follows, the straight lines collinear to v1∧vn are called horizontal, and those orthogonal to v1∧vn −→ are vertical. A non-vertical vector a is called left if its projection onto v1 ∧ vn has −−→ −→ the same direction as v1vn, and otherwise such a vector a is called right. Also, 2 we consider the line v1 ∧ vn which separates R into two half-planes, and we call the part containing v the lower half-plane, and the opposite one is called the upper half-plane. Similarly to non-vertical vectors, we split the non-horizontal vectors into two classes, which are to be called the upgoing and downgoing vectors. A point v is said to lie higher than u if the vector uv−→ goes up. The quantity xc(conv v) is also simply called the extension complexity of v.

Deﬁnition 32. Let v = (v1, . . . , vn) be a thin cw-sequence. For indexes i,ˆı, k, ,ˆ j satisfying 1 6 i < ˆı < k < ˆ < j 6 n, we deﬁne ρ(v, i,ˆı, k, ,ˆ j) as the ray passing from the point (vi ∧ vˆı) ∧ (vˆ ∧ vj) towards vk.

Lemma 33. Assume v = (v1, . . . , vn) is a thin sequence, and indexes i,ˆı, k, ,ˆ j are as in Deﬁnition 32. Then ρ(v, i,ˆı, k, ,ˆ j) ∩ conv v is a line segment s in which vk is one of the endpoints, and vk lies between the apex of ρ and the other endpoint of s. SUBLINEAR EXTENSIONS OF POLYGONS 13

Figure 5. In this example, the ray ρ(v, 1, 2, 3, 4, 5) is left upgoing, and ρ(v, 3, 5, 7, 8, 9) is vertical upgoing.

Proof. Let Φ be the set of all edge-deﬁning inequalities of conv v, and let ρ0 be the part of the ray ρ obtained by removing all the points between vk and its apex o = (vi ∧ vˆı) ∧ (vˆ ∧ vj). We write ϕ1 > 0, ϕ2 > 0 for the inequalities corresponding to the edges passing through vk, and, since v is thin, these inequalities should fail at o. So the quantities ϕ1, ϕ2 increase as we move from o towards vk, they vanish 0 at vk, and hence they should be positive on ρ . All the other inequalities in Φ are 0 strict at vk, so a small part of ρ near its beginning lies in the interior of conv v.

Deﬁnition 34. We say that the ray ρ as in Lemma 33 enters conv v at vk and leaves conv v through the other endpoint of s.

Example 35. The ray ρ(v, 1, 2, 3, 4, 5) on Figure 5 enters conv v at v3 and leaves through a point on the segment conv{v8, v9}. The ray ρ(v, 3, 5, 7, 8, 9) enters at v7 and leaves through a point on conv{v1, v9}. We finalize the section with a general property of acute diagrams whose bases have thin vertex sequences. We say that the continuation of an oriented segment uv−→ is the ray with the apex at u and direction towards v. Lemma 36. Let ∆ be an acute diagram whose base has the vertices that form a thin sequence v = (v1, . . . , vn). We consider the orientation of ∆ defined as in Remark 20 with v1 taken as a sink node. Consider an arbitrary inner node u of ∆ reachable by the oriented paths from the set Λ(u) of base nodes, and assume that vn is not in Λ(u). Let ` be a horizontal line that lies in the upper half-plane with respect to v. Assume that the continuation of any main arc coming from a vertex in Λ(u) intersects ` at some point. If the set of all such intersection points has convex hull σ, then the continuation of the arc coming from u intersects ` at a point in σ. Proof. Using the argument as in Lemma 23, we can find a base arc conv{v, w} with v, w in Λ(u) such that the main arcs passing from v, w have the common end. With an inductive application of the procedure as in Lemma 25, we can assume without loss of generality that only main arcs can have u as an endpoint, and also we can assume that there are exactly two such arcs e1, e2. In this case, it is straightforward to see that the arc coming from u lies in the cone with the apex u and extreme rays pointing towards the directions of e1, e2. 8. Good sequences We give several further auxiliary definitions to describe a situation that we call good because the corresponding sequences allow small extended formulations. 14 YAROSLAV SHITOV

Deﬁnition 37. If t, n are positive integers and G is a subset of {1, . . . , n}, then G 0 0 0 is called t-scattered if for g, g ∈ G we have either g = g or |g − g | > t. Deﬁnition 38. Assume n > 5 is an integer and G ⊂ {3, 4, . . . , n − 2} is a 3- scattered subset. If v = (v1, . . . , vn) is a thin sequence, then the G-envelope of v is the sequence vG obtained from v by replacing the points vg−1, vg, vg+1 with

og := (vg−2 ∧ vg−1) ∧ (vg+1 ∧ vg+2), for all g ∈ G.

As we see, the sequence vG is obtained from v by a removal of three vertices and an addition of one new vertex, for all g ∈ G, which means that vG contains n−2|G| vertices. More precisely, the polygon conv vG is cut by all the edge-deﬁning inequalities of conv v except 2|G| of them, namely, those corresponding to the edges conv{vg−1, vg} and conv{vg, vg+1} for g ∈ G. Further, we can see that vG is correct.

Figure 6. The G-envelope of the sequence on Figure 5 with G = {3, 7}.

Observation 39. If the sequence v as in Deﬁnition 38 is cw-ordered, then vG is a cw-ordered sequence with n − 2|G| vertices.

Proof. Since v is thin, the position of og is such that (vg−2, og, vg+2) is a cw-ordered sequence, and the general statement follows by the induction on |G|. The motivation of the following definition lies in Lemma 41 below, which is a reformulation of Theorem 28 in terms that are more comfortable to work with later. Definition 40. Assume n > 5 is an integer and G ⊂ {3, 4, . . . , n − 2} is a 3- scattered subset. A correct sequence v = (v1, . . . , vn) is called G-good if v is thin and, additionally, there is an acute diagram with the base conv vG such that, for any g ∈ G, the main edge passing from og contains vg. A particular instance of Definition 40 is Figure 7, which corresponds to n = 12 and shows a G-good sequence (v1, . . . , v12) with G = {3, 7, 10}.

Lemma 41. Let n > 5 and δ > 1 be integers, and let v = (v11, v21, . . . , vn1) be a thin sequence. Let G ⊂ {3, 4, . . . , n − 2} be a 3-scattered subset. Let V = (vgj) be an array of points in which the indexes (g, j) run over G × {1, . . . , δ}. We assume that, for any j ∈ {1, . . . , δ}, we obtain a good sequence if we replace vg1 with vgj whenever g ∈ G. Then xc(conv v ∪ V) 6 n − |G| + δ. Proof. The idea of this proof is to apply Theorem 28; we deﬁne the polygon P as in that theorem to be the convex hull of the G-envelope of v. The set S as in Theorem 28 is chosen as {og : g ∈ G}, where

og := (vg−2,1 ∧ vg−1,1) ∧ (vg+1,1 ∧ vg+2,1) SUBLINEAR EXTENSIONS OF POLYGONS 15 are the new vertices in the G-envelope of v. We pick vg−1,1 and vg+1,1 as the 0 00 choice of s , s corresponding to the vertex s = og as in Theorem 28. Since by the formulation of the lemma, a replacement of vg1 with vgj returns a good sequence, and hence, in particular, a correct sequence, we see that the vertices vg1, . . . , vgδ lie in the interior of the triangle conv{og, vg−1,1, vg+1,1}; these triangles are disjoint for diﬀerent g ∈ G again because the sequence v is correct. The penultimate sentence of Theorem 28 means in our setting that the sequence obtained by replacing vg1 with vgj is G-good, and hence it is also valid by the assumptions of the current lemma. So we have checked the conditions of Theorem 28, and we apply it to conclude that xc(conv v ∪ V) 6 (n − 2|G|) + |G| + δ = n − |G| + δ. We ﬁnalize this section with an important example of a good sequence.

Lemma 42. Assume n > 5 is an integer and G ⊂ {3, 4, . . . , n − 2} is a 3-scattered subset. Assume v = (v1, . . . , vn) is a thin sequence such that, for all g ∈ G, the ray ρ(v, g − 2, g − 1, g, g + 1, g + 2) as in Deﬁnition 32 leaves conv v through the relative interior of the edge conv{v1, vn}. Then v is G-good.

Figure 7. An instance of Lemma 42 with n = 12 and G = {3, 7, 10}.

Proof. We recall that the G-envelope of v is obtained by adding the point

og := (vg−2 ∧ vg−1) ∧ (vg+1 ∧ vg+2) and removing vg−1, vg, vg+1 for every g ∈ G. According to Definition 40, we need to construct an acute diagram ∆ in which the base is the G-envelope of u, and the main edge of ∆ passing from og contains vg for all g ∈ G. We use Corollary 26 to build an acute diagram ∆ with such a base in which (1) for any g ∈ G, the main edge from any og goes towards vg, (2) for any k∈ / {1, n, g − 1, g, g + 1}, the main edge from vk passes in the vertical direction, (3) the main edge from vn passes towards a point u0 that lies sufficiently close to the middle of the edge conv{v1, vn} in the interior of conv v. We need to show that the main edge going from og does actually contain vg. We consider the orientation of ∆ as in Remark 20 with v1 as a sink vertex. We proceed with a proof by contradiction; we assume that the desired statement is false, that is, there is an index j ∈ G for which the main edge passing from oj does 16 YAROSLAV SHITOV not contain vj. This means that some edge e of ∆ has the ending point in the interior of conv{og, vj}, and hence the ending point of e lies outside conv v. Case 1. There exists an oriented path from vn to e. This is impossible because the oriented path starting at vn reaches v1 without leaving conv v (in fact, it does not leave a sufficiently small neighborhood of conv{v1, vn} by the choice of u0). Case 2. There is no oriented path from vn to e. According to Lemma 36, the edges not having vn as a predecessor can exit conv v through the interior of conv{v1, vn} only, so any oriented path connecting a base vertex of ∆ to the endpoint of e should cross the interior of conv{v1, vn}, and hence it should cross the oriented path between vn and v1. Therefore, the situation falls into the already refuted Case 1; since these cases cover all possibilities, the proof is complete. 9. Extracting a slanted subsequence We are ready to introduce the class of so-called slanted sequences, which will be shown in Section 11 to contain sufficiently large subsequences of small extension complexity. In this section, we are going to explain how to extract a large slanted subsequence from a given sequence of large extension complexity. These two results will allow us to conclude the proof of Theorem 5 in Section 12.

Definition 43. Let v = (v1, . . . , vt) be a thin cw-sequence, β ∈ (0, π/2), δ > 0. We say that v is cw-slanted to an angle β with tolerance δ if, for all i,ˆı, ,ˆ j satisfying 1 6 i < ˆı < ˆ < j 6 t, the ray ρ(v, i,ˆı, k, ,ˆ j) as in Definition 32 satisfies −−−−→ (9.1) ∠ vˆ ∧ vj, ρ(v, i,ˆı, k, ,ˆ j) < β for all k satisfying ˆı < k < ˆ except for at most δ such values of k. Observation 44. Let k0, k00 be two indexes satisfying ˆı < k0 < k00 < ˆ in the notation of Definition 43. If (9.1) fails for k = k00, then it fails for k = k0 as well. Proof. Follows because v is correct. Observation 45. Let β, δ, v be as in Definition 43. Then any (δ+1)-scattered subsequence of v is cw-slanted to the angle β with tolerance 0. Proof. Follows from Observation 44.

Remark 46. If a correct sequence v = (v1, . . . , vt) is not cw-ordered, which means that it is ccw-ordered, then we say that v is cw-slanted with parameters as in Defi- nition 43 if the reversed sequence satisfies the assumptions of this definition. Remark 47. We define the dual concept of ccw-slantedness by replacing every occurrence of ‘cw’ with ‘ccw’ and vice versa in Definition 43 and Remark 46. The goal of the remaining part of this section is to prove Lemma 49 below; we need to begin with an important special case of it. Lemma 48. Let G ⊂ {3, 4, . . . , n − 3} be a 4-scattered subset with n > 5, and let v = (v1, . . . , vn) be a thin cw-sequence. Assume that, for any g ∈ G, the ray ρ(v, g − 2, g − 1, g, g + 1, g + 2) is left, and vg is located to the right of the point

ωg := (vg−2 ∧ vg−1) ∧ (vg+2 ∧ vg+3).

Then there exist arbitrarily high points u1 and un such that u1 lies on the ray passing from v2 towards v1, and un lies on the ray passing from vn−1 towards vn, for which the sequence (u1, v2, . . . , vn−1, un) is G-good. SUBLINEAR EXTENSIONS OF POLYGONS 17

Figure 8. An instance of Lemma 48 with n = 11 and G = {3, 8}.

Proof. For any g ∈ G, we set

og := (vg−2 ∧ vg−1) ∧ (vg+1 ∧ vg+2) and denote by `g the vertical line passing through ωg. Also, we deﬁne

αg := `g ∧ (og ∧ vg), and we set the points u1, un as in the formulation of the lemma so that the segment connecting them is horizontal and lies higher than the point αg for any g ∈ G. Proving that (u1, v2, . . . , vn−1, un) is a G-good sequence requires an acute diagram whose base is the convex hull of the G-envelope of (u1, v2, . . . , vn−1, un), so the base vertices of such a diagram can be separated into the following three types: (A) u1, un, (B) og and vg+2 with g ∈ G, (C) all other vertices. By the deﬁnition of the G-envelope, we note that the type C vertices should include those vj for which j∈ / {1, n} and also j∈ / {g − 1, g, g + 1, g + 2} for any g ∈ G. Now we are ready to construct the diagram ∆ with the use of Lemma 26. The main edge passing from u1 gets assigned the direction towards an inner point close to the middle of the segment conv{u1, un}; the vertices of type C are assigned the vertical direction of the corresponding main edges. Finally, the vertices of type B have main edges passing towards αg with g ∈ G.

Figure 9. The proof of Lemma 48 with the polygon on Figure 8.

In order to complete the proof, we need to check that the main edge passing from og does indeed contain vg for any g ∈ G. Indeed, a stronger fact follows by the induction on the cardinality of G, namely, that the two main edges passing from og and vg+2 meet at αg, and the third arc adjacent to αg goes vertically until it meets 18 YAROSLAV SHITOV the path between u1 and un close to the segment conv{u1, un}, and, similarly, a main edge passing from a vertex of type C goes in the vertical direction until it meets the path between u1 and un, as it is the case with v6 on Figure 9. Now we are ready to proceed with the main result of this section. Lemma 49. Let δ, τ, m be positive integers and n = 8τm. Let α, β be positive reals with π/2 > β > 2α. Assume v = (v1, . . . , vn) is an α-thin sequence such that any its subsequence with m points is neither cw-slanted nor ccw-slanted relative to the angle β and tolerance 2δ. Then v has a subsequence with (6 + δ)τ points with extension complexity not exceeding 6τ + δ + 1. Proof. We assume without loss of generality that the vertices of v are enumerated in the cw-order. The assumption on the lack of slanted subsequences implies that

(v1, . . . , vm), (v2m+1, . . . , v3m),..., (v(8τ−2)m+1, . . . , v(8τ−1)m) are not ccw-slanted with respect to the angle β and tolerance 2δ, and also

(vm+1, . . . , v2m), (v3m+1, . . . , v4m),..., (v(8τ−1)m+1, . . . , v8τm) are not cw-slanted relative to the same parameters. This means that, for any 0 0 0 q ∈ {0,..., 4τ − 1}, we can ﬁnd iq, ˆıq, ˆq, jq and iq, ˆıq, ˆq, jq satisfying

2qm + 1 6 iq < ˆıq < ˆq < jq 6 (2q + 1)m, 0 0 0 0 (2q + 1)m + 1 6 iq < ˆıq < ˆq < jq 6 (2q + 2)m, for which the ray ρkq := ρ(iq,ˆıq, kq, ˆq, jq) satisﬁes −−−→ (9.2) ∠ vˆıq viq , ρkq > β for all kq in a set ∆q ⊆ {ˆıq + 1,..., ˆq − 1} of cardinality at least 2δ, and the ray 0 0 0 0 0 ρ 0 := ρ(i ,ˆı , k , ˆ , j ) satisﬁes kq q q q q q −−−→ v 0 v 0 , ρ 0 β ∠ ˆq jq kq > 0 0 0 0 for all kq in a set ∆q ⊆ {ˆıq + 1,..., ˆq − 1} of cardinality at least 2δ. Since v is 0 cw-ordered, every point in ∆q lies to the right of any point of ∆q, so for the point

ω := (v ∧ v ) ∧ (v 0 ∧ v 0 ) q iq ˆıq ˆq jq we have at least one of the two options: (a) every point of ∆q lies to the right of ωq, 0 (b) every point of ∆q lies to the left of ωq. We can assume without loss of generality that the quantity of those q for which (a) holds is not less than the corresponding quantity for (b), because otherwise this property can be satisﬁed on the sequence obtained from v by the axial symmetry. So we can ﬁnd a set Q ⊂ {0,..., 4τ − 1} of cardinality 2τ such that any q ∈ Q has the property (a). Further, an index q ∈ Q is called left-admissible if there is a set

Lq ⊂ ∆q of cardinality δ consisting of those kq ∈ ∆q for which the ray ρkq is left. Otherwise, an index q ∈ Q is called right-admissible. The situation splits into two possible cases, which we need to consider separately. Case 1. Assume that the quantity of the left-admissible indexes in Q is not less than that of the right-admissible indexes. This means that we can pick a set Ql of cardinality τ such that, for any q in Ql, there exists a set Lq ⊂ ∆q of δ values of SUBLINEAR EXTENSIONS OF POLYGONS 19

kq for which the ray ρkq is left. Then we define the subsequence u of v containing the points [ n o L ∪ v , v , v , α , v 0 , q iq ˆıq ˆq q jq q∈Ql where α := (v ∧ v ) ∧ (v 0 ∧ v 0 ). q ˆq jq ˆq jq For any choice of kq ∈ Lq, the subsequence uq of u consisting of [ n o v , v , v , v , α , v 0 iq ˆıq kq ˆq q jq q∈Ql satisfies the assumptions of Lemma 48 with G being the set of the positions of the vkq ’s. According to this lemma, we see that the sequence uq becomes G-good after stretching the first and last vertices, by which we mean that one replaces these vertices by two higher points, one of which is taken on the ray passing from the second point towards the first point, and the other one is chosen on the ray passing from the penultimate point towards the last point, respectively. Also, let u0 be a sequence obtained from u by stretching the first and last vertices; this sequence has (5 + δ)τ vertices. According to Lemma 41, we can get 0 xc(conv u ) 6 6τ − τ + δ = 5τ + δ. Now we can cut conv u0 by the line joining the first and last points of u, which acts like a takeback of the stretching operation. In other words, this leaves us with conv u, which we further cut along v ∧ v 0 for all q ∈ Q , thus adding the jq ˆq l 0 v , v 0 to u and getting rid of all the points α defined above. So we get from u a jq ˆq q subsequence of v of cardinality (6 + δ)τ for the cost of τ + 1 additional cuts, each of which can worsen our bound only by at most one, according to Observation 9, so this subsequence has extension complexity not exceeding 6τ + δ + 1. Case 2. The set Q has at least as many right-admissible indexes as left-admissible indexes. Similarly to the consideration in the first paragraph of Case 1, we find a set Qr of cardinality τ such that, for any q in Qr, there exists a set Rq ⊂ ∆q of

δ values of kq for which the ray ρkq is not left. We deﬁne the subsequence w of v containing the points [ Rq ∪ viq , vˆıq , vˆq , vjq ,

q∈Qr and we denote by wr, wl the ﬁrst and last points of w, respectively. According to Observation 6, we have −−−→ −−−→ −−−→ −−−→ ∠ ρkq , wrwl > ∠ ρkq , vˆıq viq − ∠ vˆıq viq , wrwl , and using the bound of (9.2) for the ﬁrst summand and the α-thinness for the second summand, we get −−−→ ∠ ρkq , wrwl > β − α > α, so for kq ∈ Rq, the ray ρkq has to leave conv w through the interior of conv{wl, wr}. According to Lemma 42, the sequence [ viq , vˆıq , vkq , vˆq , vjq

q∈Qr is G-good for any choice of kq ∈ Rq, where G is the set of the positions of the vkq ’s. Using Lemma 41, we conclude that the extension complexity of w is at most 20 YAROSLAV SHITOV

5τ − τ + δ = 4τ + δ. Since the cardinality of w is (4 + δ)τ, we can add 2τ arbitrary points of v to w, and the resulting sequence will have extension complexity not exceeding 6τ + δ according to Lemma 8. 10. Decreasing slanted sequences In the previous section, we introduced the notion of a slanted sequence. Roughly speaking, we showed that a polygon of large extension complexity should admit a sufficiently large slanted subsequence. This result allows us to concentrate on slanted sequences, and the following definition turns out to be useful. We recall that the notation dist(A, B) stands for the distance between two sets A, B ⊂ R2. 0 0 0 Definition 50. Let u = (u1, u1, w12, u2, u2, w23, . . . , ws−1,s, us, us) be a thin cw- sequence with 3s − 1 points. We say that u is cw-decreasing if the inequality 0 0 dist(wq−1,q, uq ∧ uq) > dist(wq,q+1, uq ∧ uq) holds for any q ∈ {2, 3, . . . , s}.

Figure 10. An instance of Deﬁnition 42 with s = 4: Among two consecutive punctured points wq−1 q and wq q+1, the left one is 0 closer to the straight line uq ∧ uq containing the thick edge.

Now we are going to show that cw-decreasing sequences admit acute extensions under an additional mild assumption on the positions of the vertices. The proof of the following lemma uses an idea similar to that of Lemma 42. 0 0 Lemma 51. Let u = (u1, u1, w12, . . . , ws−1,s, us, us) be a thin cw-decreasing sequence written in the cw-order. Assume that the ray passing from 0 0 ωq,q+1 := (uq ∧ uq) ∧ (uq+1 ∧ uq+1) towards wq,q+1 is left, for any q ∈ {1, . . . , s − 1}. Then u is G-good, where G = {3, 6,..., 3s − 3} is the set of the positions of the wq,q+1’s. Proof. To prove the desired result, we need to construct an acute diagram ∆ whose base is the G-envelope of u, and the main edge of ∆ passing from every ωq,q+1 should contain wq,q+1. According to Deﬁnition 40, the G-envelope of u is 0 (u1, w12, w23, . . . , ws−1,s, us), and, as in the proof of Lemma 42, we use Corollary 26 to build an acute diagram ∆ in which (1) for any q ∈ {1, . . . , s−1}, the main edge from any ωq,q+1 goes towards wq,q+1, 0 (2) the main edge from us passes towards a point u0 that lies suﬃciently close 0 to the middle of the edge conv{u1, us} in the interior of conv u. We need to show that the main edge going from ωq,q+1 does actually contain wq,q+1. We consider the orientation of ∆ as in Remark 20 with u1 as a sink vertex. SUBLINEAR EXTENSIONS OF POLYGONS 21

We proceed with a proof by contradiction; we assume that the desired statement is false, that is, there is an index j ∈ {1, . . . , s − 1} for which the main edge passing from ωj,j+1 does not contain wj,j+1. This means that some edge e of ∆ has the ending point ε in the interior of conv{ωj,j+1, wj,j+1}, and hence this point ε lies outside conv u. Furthermore, we assume that j is the minimal index that possesses this property. 0 Case 1. There exists an oriented path from us to e. Similarly to the proof of 0 Lemma 42, this is impossible because the oriented path from us to u1 lies suﬃciently 0 close to conv{u1, us}. Case 2. There exists an oriented path from some ωt t+1 to e with t > j. This is again impossible because wj j+1 is located to the right of ωt t+1, and every edge of 0 ∆ that is not reachable from us is left because of Lemma 36. Since Cases 1 and 2 are invalid, we can use Lemma 22 and conclude that ε is reachable by an oriented path from ωj−1,j. We denote this path by π, and we conclude, by the minimality of j, that the main edge of ωj−1 j passes through wj−1 j, which means that wj−1 j lies on π. Using the thinness of u, we conclude that the 0 distance from a point x ∈ π to uj ∧ uj grows as we move x towards the direction of π. Since u is cw-decreasing, we have 0 0 dist(wj,j+1, uj ∧ uj) < dist(wj−1,j, uj ∧ uj), so the point wj,j+1 occurs on π earlier than wj−1,j, which is a contradiction. The main result of this section is a corollary of Lemma 51.

Corollary 52. Let δ, τ be positive integers; let α, β be positive reals with α + β 6 π/2. Suppose n = 4δτ + 2 and let v = (v1, . . . , vn) be an α-thin cw-sequence. If v is cw-slanted to the angle β with tolerance 2δ, and if the subsequence 0 v = (v1, v2, v4δ, v4δ+1, v4δ+2, v8δ, . . . , vn−1, vn) is cw-decreasing, then v possesses a subsequence of 2δτ + 2 points with extension complexity not exceeding 2τ + 2δ. Proof. According to Deﬁnition 43, every choice of the index q = {0, . . . , τ − 1} allows a set ∆q ⊂ {4qδ + 3,..., 4qδ + 4δ} of cardinality 2δ − 2 such that the ray

ρkq := ρ(v, 4qδ + 1, 4qδ + 2, kq, 4(q + 1)δ + 1, 4(q + 1)δ + 2) as in Deﬁnition 32 satisﬁes −−−−−−−−−−−−−−→ ∠(ρkq , v4(q+1)δ+1v4(q+1)δ+2) < β −−→ for kq ∈ ∆q. According to Observation 6, the angle ∠(ρkq , v1vn) does not exceed −−−−−−−−−−−−−−→ −−→ −−−−−−−−−−−−−−→ ∠(ρkq , v4(q+1)δ+1v4(q+1)δ+2) + ∠(v1vn, v4(q+1)δ+1v4(q+1)δ+2) < β + α 6 π/2, so the ray ρkq is left for any kq ∈ ∆q. We also have

(10.1) dist(vkq , v4(q+1)δ+1 ∧ v4(q+1)δ+2) > dist(v4(q+1)δ, v4(q+1)δ+1 ∧ v4(q+1)δ+2),

(10.2) dist(v4(q+2)δ, v4(q+1)δ+1 ∧ v4(q+1)δ+2) > dist(vkq+1 , v4(q+1)δ+1 ∧ v4(q+1)δ+2) for all kq ∈ ∆q and kq+1 ∈ ∆q+1 because v is correct, and also

dist(v4(q+1)δ, v4(q+1)δ+1 ∧ v4(q+1)δ+2) > dist(v4(q+2)δ, v4(q+1)δ+1 ∧ v4(q+1)δ+2) 22 YAROSLAV SHITOV because v0 is cw-decreasing. Putting the last inequality together with (10.1)–(10.2), we get

dist(vkq , v4(q+1)δ+1 ∧ v4(q+1)δ+2) > dist(vkq+1 , v4(q+1)δ+1 ∧ v4(q+1)δ+2), which means that the sequence (v1, v2, vk0 , v4δ+1, v4δ+2, vk1 , . . . , vn−1, vn) is cw- decreasing for any choice of vkq ∈ ∆q, and by Lemma 51 this sequence is G-good, where G is the set of the positions of the vkq ’s. It remains to apply Lemma 41 to see that the subsequence formed by the set

{v1, v2, v4δ+1, v4δ+2, . . . , vn−1, vn} ∪ ∆0 ... ∪ ∆τ−1 has extension complexity not exceeding (3τ + 2) − τ + (2δ − 2) = 2τ + 2δ. 11. Extensions for slanted sequences In this section, we ﬁnalize our discussion of slanted sequences by proving a general result concerning their extension complexities. The following auxiliary lemma is quite straightforward.

Lemma 53. Let u = (u2, u3, u4, u5, u6) be a thin cw-ordered sequence. Assume that ρ is a ray with apex at u4. If ρ intersects conv{u5, u6}, then −−→ −−→ −−→ ∠ (u2u3, ρ) < ∠ (u2u3, u5u6) .

Proof. Let w6 be the intersection point of conv{u5, w6} and ρ. We get −−→ ∠ (u2u3, ρ) = ∠u2u3u4 + ∠u3u4w6, −−→ −−−→ ∠ (u2u3, u5w6) = ∠u2u3u4 + ∠u3u4u5 + ∠u4u5w6, and we are done because ∠u3u4u5 + ∠u4u5w6 > ∠u3u4w6. The following two lemmas are based on direct computations. In Lemma 54 below, it is useful for us to develop a representation of a thin sequence in Cartesian coordinates diﬀerent from that of Deﬁnition 31. Namely, for a given thin sequence v = (v2, . . . , vt) with t > 4 that is written in the cw-order, we can construct a congruence transformation of R2 that sends (i) v2 to a point (a, 0) with a > 0, v3 to the point (0, 0), (ii) v4, . . . , vt to points with positive second coordinates. In fact, the images of two distinct points as in (i) determine the congruence transformation up to a symmetry, and since our sequence corresponds to a convex polygon, the remaining points should have second coordinates of the same sign.

Figure 11. The coordinate system as in the proof of Lemma 54. SUBLINEAR EXTENSIONS OF POLYGONS 23

Lemma 54. Let v = (v1, . . . , vt) be an α-thin sequence cw-slanted to an angle β with tolerance 0, where α + β 6 π/2. Then, for any tuple of indexes i,ˆı, k, ,ˆ j satisfying 1 6 i < ˆı < k < ˆ < j 6 t, we have dist(vˆı, vk) sin(∠vivˆıvk) > dist(vk, vˆ) sin(∠vkvˆvj).

Proof. We introduce the notation Lpq = dist(vp, vq), apqr = π − ∠vpvqvr and as = as−1,s,s+1, and also we write ρiˆıkjˆ = ρ(v, i,ˆı, k, ,ˆ j). We assume without loss of generality that v is written in the cw-order, and also that i = 2, ˆı = 3, k = 4, ˆ = 5, j = 6. The congruence transformation described above allows us to take

v2 = (L23, 0), v3 = (0, 0), v4 = (−L34 cos a3,L34 sin a3),

v5 = v4 + (−L45 cos(a3 + a4),L45 sin(a3 + a4)),

v6 = v5 + (−L56 cos(a3 + a4 + a5),L56 sin(a3 + a4 + a5)). −−→ Now we want to compute the angle between the ray ρ23456 and the vector v2v3, which goes towards the negative direction of the x-axis. This ray has the apex at o := (v2 ∧ v3) ∧ (v5 ∧ v6) and its direction is towards v4; a direct computation gives −L sin a − L sin(a + a ) o = 45 5 34 4 5 , 0 . sin(a3 + a4 + a5) −−→ The tan of the angle between ρ23456 and v2v3 equals L sin a sin(a + a + a ) (11.1) 34 3 3 4 5 , L34 sin a3 cos(a3 + a4 + a5) − L45 sin a5 and we are done because the denominator of (11.1) should be positive (if this was −−→ −−→ not the case, we would have ∠(ρ23456, v2v3) > π/2 and hence ∠(ρ23456, v5v6) > π/2 − α > β, which is a contradiction to the cw-slantedness). We proceed with a deeper result based on computations similar to Lemma 54.

Lemma 55. Let v = (v1, . . . , v10) be a π/6-thin sequence cw-slanted relative to the angle π/3 with tolerance 0. We assume that γ3 > 1, γ4 > 1, γ5 > 1, γ6 > 1, where

dist(vτ+3, vτ+1 ∧ vτ+2) (11.2) γτ = . dist(vτ , vτ+1 ∧ vτ+2) Then there are indexes i,ˆı, k, ,ˆ j satisfying 2 6 i < ˆı < k < ˆ < j 6 9 such that ρ(v, i,ˆı, k, ,ˆ j) leaves conv v through the relative interior of conv{v1, v10}. Proof. We continue to use the notation of the proof of Lemma 54. As we saw from the cw-slantedness, the ray ρ23456 should be left, so the only possibility for it to fail the desired statement, that is, not to leave conv v through the relative interior of conv{v1, v10}, is that ρ23456 intersects conv{v6, v10}. Using Lemma 53 with v2, v3, v4, v6, v10 taking the roles of u2, u3, u4, u5, u6, respectively, and with the part of ρ23456 beginning at v4 in the role of ρ, we get −−→ −−−→ −−→ (11.3) ∠(ρ23456, v2v3) < ∠(v6v10, v2v3).

We note that the right-hand side of (11.3) is a3 +a4 +a5 +a5 6 10, and the left-hand side of (11.3) was computed in the proof of Lemma 54 and equals the arctan of the expression (11.1). We get L34 sin a3 sin(a3 + a4 + a5) (11.4) arctan − a3 − a4 − a5 < a5 6 10. L34 sin a3 cos(a3 + a4 + a5) − L45 sin a5 24 YAROSLAV SHITOV

We have L56 sin a5 6 10 < L45 sin a4 by the result of of Lemma 54, so the right- hand side of (11.4) can be replaced by arcsin(L45 sin a4/L56). We do this and further take the tan function of both sides of (11.4); we√ recall that tan(x + y) = (tan x + tan y)/(1 − tan x tan y) and tan(arcsin x) = x/ 1 − x2. We get

L45 sin a5 sin(a3 + a4 + a5) L45 sin a4 < q , L34 sin a3 − L45 sin a5 cos(a3 + a4 + a5) 2 2 2 L56 − L45 sin a4 and now we are able to remove the second listed summand of the denominator of the left-hand side. Further elementary transformations yield q 2 2 2 sin a5 sin(a3 + a4 + a5) L56 − L45 sin a4 < L34 sin a3 sin a4.

We note that L45 sin a5 < L34 sin a3 by the result of Lemma 54, which gives s 2 2 2 2 L34 sin a4 sin a3 (11.5) sin a5 sin(a3 + a4 + a5) L56 − 2 < L34 sin a3 sin a4, sin a5 and now we use the equation (11.2) with τ = 3, which means that

L56 sin a5 L34 sin a4 = , γ3 and we make the replacement of L34 sin a4 in (11.5) to get s 2 2 2 L56 sin a3 L56 sin a3 sin a5 sin a5 sin(a3 + a4 + a5) L56 − 2 < , γ3 γ3 and further elementary cancellations give q 2 2 (11.6) sin(a3 + a4 + a5) γ3 − sin a3 < sin a3. This inequality is obviously false if the square root is greater than or equal to one, p 2 so we have γ3 < 1 + sin a3, which implies

(11.7) γ3 < 1.5.

Taking into account the inequality γ3 > 1 from the assumptions of the lemma and the inequality a3 < π/6 that comes from the π/6-thinness of v, we transform (11.6) into p 2 sin(a3 + a4 + a5) 1 − (sin π/6) < sin a3, from which it can be deduced that

(11.8) a4 < a3 by standard tools of calculus. Now we apply similar considerations with the rays ρ56789 and ρ23458 in the role of the ray ρ23456 above. In the above consideration, the only case of the inequality γi > 3 that we used was i = 3, so the ability to make the same computations for ρ56789 is justiﬁed by the analogous inequality γ6 > 1 in the formulation. Similarly, the case of ρ23458 is guaranteed by

dist(v8, v4 ∧ v5) dist(v6, v4 ∧ v5) > = γ3 > 1. dist(v3, v4 ∧ v5) dist(v3, v4 ∧ v5) SUBLINEAR EXTENSIONS OF POLYGONS 25

In particular, if the ray ρ56789 was not leaving conv v through the relative interior of conv{v1, v10}, we would get

(11.9) a7 < a6 as the analogue of (11.8). Similarly, the analogue of (11.7) for ρ23458 reads dist(v , v ∧ v ) 8 4 5 < 1.5 dist(v3, v4 ∧ v5) or L56 sin a5 + L67 sin(a5 + a6) + L78 sin(a5 + a6 + a7) < 1.5L34 sin a4, which implies

(11.10) (L56 + L78) sin a5 < 1.5L34 sin a4.

Similarly, the assumption γ5 > 1 in the formulation of the lemma means that

dist(v8, v6 ∧ v7) > 1 dist(v5, v6 ∧ v7) or L78 sin a7 > L56 sin a6. Together with (11.9), this inequality implies L78 > L56, and we further apply (11.10) and get 2L56 sin a5 < 1.5L34 sin a4, or

2 dist(v6, v4 ∧ v5) < 1.5 dist(v3, v4 ∧ v5), which means that γ3 < 3/4 and contradicts to the assumption γ3 > 1.

Deﬁnition 56. A correct sequence (v2, v3, . . . , vk) is called perfect if the quantity γτ as in (11.2) satisﬁes γτ > 1 for all τ ∈ {3, 4, . . . , k − 3}. We proceed to the main result on the extension complexity of slanted sequences.

2 Theorem 57. Let δ > 1 be an integer and n = 8δ . Let v = (v0, . . . , vn) be a π/6-thin sequence that is cw-slanted to the angle π/3 with tolerance 2δ. Then v has a subsequence of size at least 0.25δ2 and extension complexity at most 3δ. Proof. Assuming without loss of generality that v is written in the cw-order, we 0 0 0 build a cw-decreasing subsequence u = (u0, u0, w01, u1, u1, w12, . . . , uh, uh) of v. We 0 take i0 = 0, u0 = v0, u0 = v4δ, w01 = v8δ, and for any q > 0, we inductively deﬁne 0 uq = v4δiq , uq = v4δ(iq +1), wq q+1 = v4δ(iq +2), where iq ∈ {iq−1 + 3, iq−1 + 4,..., 2δ − 4} is the smallest index such that

dist(v4δ(iq +2), v4δiq ∧ v4δ(iq +1)) < dist(wq−1 q, v4δiq ∧ v4δ(iq +1)), 0 and if no such iq exists, then we take uq = vn−4δ, uq = vn, and we set h = q. The situation splits into the two cases, which we need to treat separately. Case 1. Assume h > 0.125δ. According to our definitions, the sequence u is cw-decreasing, so we can apply Corollary 52 with the subsequence 0 0 0 (u0, u0, w01, u1, u1, w12, . . . , uτ , uτ ) in the role of v0, where we can take τ = dδ/8e because of the initial assumption 2 of Case 1. This gives a subsequence of v with at least 2δτ > 0.25δ vertices and extension complexity not exceeding 2δ + 2τ < 3δ. Case 2. Assume h < 0.125δ. For any q ∈ {0, . . . , h − 1}, we define the sequence σq := v4δ(iq +2), v4δ(iq +3), . . . , v4δ(iq+1+1) , 26 YAROSLAV SHITOV which consists, in other words, of those points of the form v4δZ that lie between 0 wq q+1 and uq+1. According to the choice of iq above, the sequence σq is perfect in the sense of Definition 56. For trivial reasons, every such σq contains |σ | m := q q 8 disjoint subsequences of eight consecutive vertices each. Any such subsequence is to be called main, and every main subsequence is perfect because every σq is perfect. The total number of main subsequences is m := m0 + ... + mh−1, and since they are disjoint, we have m 6 (2δ − 1)/8 < 0.25δ. We also have h−1 h−1 h−1 ! X |σq| X |σq| 7 1 X 7h 2δ − 1 7h m = − = |σ | − = − , 8 > 8 8 8 q 8 8 8 q=0 q=0 q=0 which implies m > 0.25δ − h. According to Lemma 55, every main subsequence contains a further subsequence µ = (µ1, µ2, µ3, µ4, µ5) such that the ray ρ(µ, 1, 2, 3, 4, 5) leaves conv v through the interior of conv{v0, vn}. According to Definition 43 and Observation 44, the ray 0 0 ρ(µ , 1, 2, 3 , 4, 5) leaves conv v through the interior of conv{v0, vn} not only for µ30 = µ3 but also for at least 2δ − 1 additional choices of a point µ30 between µ2 0 and µ3 on the sequence v, where µ = (µ1, µ2, µ30 , µ4, µ5) denotes the sequence obtained from µ by replacing µ3 with a such vertex µ30 . According to Lemma 42, we obtain a G-good sequence if we take the concatenation of all (µ1, µ2, µ30 , µ4, µ5) in which µ30 is arbitrarily chosen from the 2δ vertices mentioned above (with G being the set of the positions of the µ30 ’s). Finally, we set U to be the sequence whose vertex set is the union of {µ1, µ2, µ4, µ5} together with all possible choices of µ30 over all sequences µ as above. We see that the sequence U has 2 2 2 4m + 2δm > (δ − 4h) + 0.5δ − 2δh > (δ − 4 · 0.125δ) + 0.5δ − 2δ · 0.125δ > 0.25δ vertices, and, according to Lemma 41, the extension complexity of U does not exceed 5m − m + 2δ 6 3δ.

12. Completing the proof In this section, we put our technical results together and complete the proof of Theorem 5. First, we prove that a π/6-thin sequence admits a suﬃciently large subsequence with small extension complexity. Theorem 58. Let v be a sequence with n = 1024τ 3 + 8τ vertices, where τ is a positive integer. If v is π/6-thin, then v contains a subsequence with at least 4τ 2 vertices and extension complexity at most 12τ. Proof. We apply Lemma 49, assuming β = π/3, δ = 4τ, and m = 8δ2 + 1. Case 1. The assertion of Lemma 49 holds. Then v contains a subsequence with (6 + δ)τ = 4τ 2 + 6τ vertices and extension complexity not exceeding 6τ + δ + 1 = 10τ + 1, which satisﬁes the bounds desired in the current theorem. Case 2. The assumption of Lemma 49 does not hold, which means that v admits a subsequence u = (u1, . . . , um) which is either cw-slanted or ccw-slanted to the angle π/3 with tolerance 2δ. According to Theorem 57, such a sequence u has a subsequence of at least 0.25δ2 = 4τ 2 points and extension complexity at most 3δ = 12τ. These bounds match the desired conclusion as well. SUBLINEAR EXTENSIONS OF POLYGONS 27

Corollary 59. Let v be a sequence with n > 263 000 vertices. If v is π/6-thin, then it contains a subsequence u with at least √ 3 n2/36 vertices with extension complexity not exceeding p3 72n/43. Proof. We take r n τ = 3 1032 and apply Theorem 58 to arbitrary subsequence of v containing 1024τ 3 +8τ points. So we are able to extract a subsequence of v with extension complexity at most r n r n r72n 12τ = 12 3 12 3 = 3 1032 6 1032 43 and containing at least 2 2 √ r n r n 3 n2 4τ 2 = 4 3 > 4 3 − 1 > 1032 1032 36 vertices, where the last inequality holds in the desired range of n > 263 000. Corollary 60. Let v be a correct sequence with n vertices. If v is π/6-thin, then its extension complexity does not exceed 324p3 n2/129.

Proof. For n 6 263 000, the result is trivial because the desired bound is greater than n. For n > 263 000, we proceed by induction and extract a subsequence u as in Corollary 59. So we see that the sequence v splits into the disjoint union of two subsequences, (1) one of which is u, and it has extension complexity not exceeding p3 72n/43; (2) the other one, w, consists of all those vertices in v that are√ not in u. The bound in Corollary 59 guarantees that w has at most n − 3 n2/36 vertices; by the inductive hypotesis we have

v √ 2 u 3 2 ! 324 u3 n xc(conv w) √ t n − , 6 3 129 36 which gives together with Lemma 8

v √ 2 u 3 2 ! r 324 u3 n 72n 324n xc(conv v) √ t n − + 3 < √ , 6 3 3 129 36 43 129n where the last inequality is valid for any positive integer n. Now we can prove Theorem 5. According to Observation 30, the vertex set of any convex n-gon P can be respresented as a disjoint union of twelve subsets V1,...,V12 such that any subset with |Vi| > 2 can be written as a π/6-thin sequence. We assume without loss of generality that each of the ﬁrst k subsets V1,...,Vk have cardinality greater than 2, and each of the remaining subsets has cardinality at most 2. According to Lemma 8, we have

xc(P ) 6 xc(conv V1) + ... + xc(conv Vk) + |Vk+1| + ... + |V12|, 28 YAROSLAV SHITOV and, denoting |Vi| by ni, we apply Corollary 60 and obtain 12 ! 324 X q (12.1) xc(P ) √ · 3 n2 6 3 i 129 i=1 √ 3 2 with n1 +...+n12 = n. Since t → t is a concave function on t > 0, the right-hand side of (12.1) attains its maximum if n1 = ... = n12 = n/12. This gives

r 2 ! 324 n √3 xc(P ) √ · 12 3 < 147 n2, 6 3 129 122 which is the desired bound.

13. Concluding remarks 2/3 We proved that pc(n) 6 147 n , that is, every convex n-gon can be obtained as a linear projection of a higher-dimensional polytope with at most 147 n2/3 facets. As outlined in Section 2, this refutes expectations expressed in recent literature that pc(n) should be close to a linear function. In view of our result, a weaker conjecture on lower bounds of polygon complexity could be proposed. √ Conjecture 61. One has pc(n) = n · α(n) with unbounded α(n). Besides from being an adaptation of previous expectations, this conjecture has some further supporting evidence. We recall a result of Padrol [43] stating that √ (13.1) wcc(d, n) > 2 dn − d − d + 1, where wcc(d, n) is the largest possible extension complexity of a polytope with n vertices in√ a d-dimensional space. If Conjecture 61 were false, we would have pc(n) ∈ O( n), which would mean that the bound (13.1) becomes asymptotically optimal when restricted to the case d = 2. This is not quite expected, because√ even for slowly growing d the function wcc(d, n) can grow much faster than dn. In fact, a known result on the so-called correlation polytope [35] gives 2 m m (13.2) wcc m , 2 > 1.5 , and the authors of [56] make a further conjecture that implies wcc m2, 2m = m 2 for any positive integer m. The√ application of (13.1) gives a bound similar to (13.2) but with 1.5 replaced by 2, which is much weaker in the asymptotical sense. As a further piece of evidence towards Conjecture 61,√ we note that the method presented in this paper does not seem to allow an O( n) upper bound for the worst-case n-gon complexity, but we were able to construct polygons whose optimal extended formulations require the construction as in Theorem 28 already with n = 9. However, as it is the case with many questions on lower bounds in combinatorial optimization, a proof of Conjecture 61 remains elusive.

14. Acknowledgments My interest to Question 1 developed after a fruitful workshop on Communication complexity, Linear optimization, and Lower bounds for the nonnegative rank of matrices, which took place at Schloss Dagstuhl in February, 2013. I would like to thank Nicolas Gillis and Fran¸coisGlineur for personal discussions on the topic and all other participants for the information that I learned on Question 1 and similar issues from their talks and open problem sessions. The ﬁrst version of this SUBLINEAR EXTENSIONS OF POLYGONS 29 paper [52], which contained a proof that pc(n) is o(n), came in 2014 as a result of the eﬀort I spent after this workshop, but the paper was not published. I would like to thank the organizers and participants of another workshop on Limitations of Convex Programming: Lower Bounds on Extended Formulations and Factorization Ranks hosted by Schloss Dagstuhl in 2015, and, in particular, I am grateful to Arnau Padrol Sureda for pointing me to [44] and a further discussion on the topic of Question 1. Also, I would like to thank Nicolas Gillis and Arnaud Vandaele for a further interesting discussion that held during my visit to the University of Mons in 2017, and I am grateful to Nicolas for invitation.

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