On Planar Greedy Drawings of 3-Connected Planar Graphs

A graph drawing is $\textit{greedy}$ if, for every ordered pair of vertices $(x,y)$, there is a path from $x$ to $y$ such that the Euclidean distance to $y$ decreases monotonically at every vertex of the path. Greedy drawings support a simple geometric routing scheme, in which any node that has to send a packet to a destination"greedily"forwards the packet to any neighbor that is closer to the destination than itself, according to the Euclidean distance in the drawing. In a greedy drawing such a neighbor always exists and hence this routing scheme is guaranteed to succeed. In 2004 Papadimitriou and Ratajczak stated two conjectures related to greedy drawings. The $\textit{greedy embedding conjecture}$ states that every $3$-connected planar graph admits a greedy drawing. The $\textit{convex greedy embedding conjecture}$ asserts that every $3$-connected planar graph admits a planar greedy drawing in which the faces are delimited by convex polygons. In 2008 the greedy embedding conjecture was settled in the positive by Leighton and Moitra. In this paper we prove that every $3$-connected planar graph admits a $\textit{planar}$ greedy drawing. Apart from being a strengthening of Leighton and Moitra's result, this theorem constitutes a natural intermediate step towards a proof of the convex greedy embedding conjecture.


Introduction
Geographic routing is a family of routing protocols for ad-hoc networks, which are networks with no fixed infrastructure -such as routers or access points -and with dynamic topology [17,30,31]. In a geographic routing scheme each node of the network actively sends, forwards, and receives packets; further, it does so by only relying on the knowledge of its own geographic coordinates, of those of its neighbors, and of those of the packet destination. Greedy routing -originally called Cartesian routing [16] -is the simplest and most renowned geographic routing scheme. In this protocol, a node that has to send a packet simply forwards it to any neighbor that is closer -according to the Euclidean distance -to the destination than itself. The greedy routing scheme might fail to deliver packets because

On Planar Greedy Drawings of 3-Connected Planar Graphs
of the presence of a void in the network; this is a node with no neighbor closer to the destination than itself. For this reason, several variations of the greedy routing scheme have been proposed; see, e.g., [8,21,22]. Apart from its failure in the presence of voids, the greedy routing protocol has two disadvantages which limit its applicability. First, in order for the protocol to work, each node of the network has to be equipped with a GPS, which might be expensive and might consume excessive energy. Second, two nodes that are close geographically might be unable to communicate with each other because of the presence of topological obstructions. Rao et al. [29] introduced the following brilliant idea for extending the applicability of geographic routing in order to overcome the above issues. Suppose that a network topology is known; then one can assign virtual coordinates to the nodes and use these coordinates instead of the geographic locations of the nodes in the greedy routing protocol. The virtual coordinates can then be chosen so that the greedy routing protocol is guaranteed to succeed.
Computing the virtual coordinate assignment for the nodes of a network corresponds to the following graph drawing problem: Given a graph G, construct a greedy drawing of G, that is a drawing in the plane such that, for any ordered pair of vertices (x, y), there exists a neighbor of x in G that is closer -in terms of Euclidean distance -to y than x. Equivalently, a greedy drawing of G is such that, for any ordered pair of vertices (x, y), there exists a distance-decreasing path from x to y, that is, a path (u 1 , u 2 , . . . , u m ) in G such that x = u 1 , y = u m , and the Euclidean distance between u i+1 and u m is smaller than the one between u i and u m , for any i = 1, 2, . . . , m − 2.
Greedy drawings experienced a dramatical surge of popularity in the theory community in 2004, when Papadimitriou and Ratajczak [27] proposed the following two conjectures about greedy drawings of 3-connected planar graphs.  Papadimitriou and Ratajczak [27,28] provided several reasons why 3-connected planar graphs are central to the study of greedy drawings. First, there exist non-3-connected planar graphs and 3-connected non-planar graphs that do not admit any greedy drawing. Thus, the 3-connected planar graphs form the largest class of graphs that might admit a greedy drawing, in a sense. Second, all the graphs with no K 3,3 -minor admit a 3-connected planar spanning graph, hence they admit a greedy drawing, provided the truth of the greedy embedding conjecture. Third, the preliminary study of Papadimitriou and Ratajczak [27,28] provided evidence for the mathematical depth of their conjectures.
In 2008 Leighton and Moitra [23,24] settled the greedy embedding conjecture in the affirmative; the same result was established (independently and slightly later) by Angelini et al. [4,5]. In this paper we show the following result.

Theorem 1. Every 3-connected planar graph admits a planar greedy drawing.
Given a 3-connected planar graph G, both the algorithm by Leighton and Moitra [23,24] and the one by Angelini et al. [4,5] find a certain spanning subgraph S of G and construct a (planar) greedy drawing of S; then they embed the edges of G not in S as straightline segments obtaining a, in general, non-planar greedy drawing of G. Thus, Theorem 1 v u G β uv (G) τ uv (G) Figure 1 The paths τuv(G) and βuv(G) in a 2-connected plane graph G.
Geometry. In this paper every angle is measured in radians, even when not explicitly stated. The slope of a half-line is defined as follows. Denote by p the starting point of and let be the vertical half-line starting at p and directed towards decreasing y-coordinates. Then the slope of is the angle spanned by a counter-clockwise rotation around p bringing to coincide with , minus π 2 . Note that, because of this definition, the slope of any half-line is assumed to be between -π 2 (included) and 3π 2 (excluded); in the following, there will be very few exceptions to this assumption, which will be however evident from the text. Every angle expressed as arctan(·) is assumed to be between -π 2 and π 2 . We define the slope of an edge uv in a graph drawing as the slope of the half-line from u through v. Note that the slope of an edge uv is equal to the slope of the edge vu plus or minus π. For a directed line , we let its slope be equal to the slope of any half-line starting at a point of and directed as . We denote by ∆pqr a triangle with vertices p, q, r, and we denote by pqr the angle of ∆pqr incident to q; note that pqr is between 0 and π. Let Γ be a drawing of a graph G and let u, v be vertices in V (G). We denote by d(Γ, uv) the Euclidean distance between u and v in Γ. We also denote by d H (Γ, uv) the horizontal distance between u and v in Γ, that is, the absolute value of the difference between the xcoordinates of u and v in Γ; the vertical distance d V (Γ, uv) between u and v in Γ is defined analogously. With a slight abuse of notation, we will use d(Γ, pq), d H (Γ, pq), and d V (Γ, pq) even if p and q are points in the plane (and not necessarily vertices of G). A drawing of a graph is a straight-line drawing if each edge is represented by a straight-line segment.
The following lemma argues that the planarity and the greediness of a drawing are not lost as a consequence of any sufficiently small perturbation of the vertex positions. s m (s M ) be the minimum (maximum, respectively) slope of an edge u i u i+1 or v j v j+1 . If s 3 − π < s m ≤ s M < s 1 (if s 3 < s m ≤ s M < s 1 + π), then Γ lies entirely to the right (to the left, respectively) of 2 , except for the vertex u.

On Planar Greedy Drawings of 3-Connected Planar Graphs
Proof. We only prove that, if s 3 − π < s m ≤ s M < s 1 , then Γ lies entirely to the right of 2 , except for the vertex u; the proof that, if s 3 < s m ≤ s M < s 1 + π, then Γ lies entirely to the left of 2 , except for the vertex u, is symmetric.
Further, it suffices to prove that the paths (u = u 1 , u 2 , . . . , u p ) and (u = v 1 , v 2 , . . . , v q ) lie to the right of 2 , except for the vertex u; indeed, if that is the case, then the planarity of Γ implies that the entire drawing Γ, except for the vertex u, lies to the right of 2 .
We now prove that the path (u = u 1 , u 2 , . . . , u p ) lies to the right of 2 , except for the vertex u; the proof for the path (u = v 1 , v 2 , . . . , v q ) is analogous.
For i = 1, . . . , p, let W i be the open wedge delimited by the half-lines starting at u i with slopes s 3 − π and s 1 ; that is, W i is the region of the plane that is spanned by a half-line starting at u i with slope s 3 − π while rotating counter-clockwise around u i until it has slope s 1 . We claim that W i contains the path (u i , u i+1 , . . . , u p ) in its interior, except for the vertex u i which is on the boundary of W i . Observe that the claim (with i = 1) implies the lemma, since W i lies to the right of 2 , given that s 2 − π ≤ s 3 − π and s 1 ≤ s 2 .
We now prove the claim by reverse induction on i. The case i = p is trivial. Hence, assume that W i+1 contains the path (u i+1 , u i+2 , . . . , u p ) in its interior, except for the vertex u i+1 which is on the boundary of W i+1 . See Fig. 2b. Since s 3 − π < s m ≤ s M < s 1 , the edge u i u i+1 lies in the interior of the wedge W i , except for the vertex u i which is on the boundary of W i . Further, since u i+1 lies in the interior of W i , the entire wedge W i+1 , and hence the path (u i+1 , u i+2 , . . . , u p ), lies in the interior of W i . This completes the induction and hence concludes the proof of the lemma.
(c) if edge uv exists, then it coincides with the path τ uv (G); and (d) for every 2-cut {a, b} of G we have that a and b are external vertices of G and at least one of them is an internal vertex of the path β uv (G); further, every non-trivial {a, b}component of G contains an external vertex of G different from a and b.
Several problems are more easily solved on (strong) circuit graphs than on 3-connected planar graphs. This is because the (strong) circuit graphs can be easily decomposed into smaller (strong) circuit graphs, and hence are suitable for inductive proofs. We now present a structural decomposition for strong circuit graphs whose main ideas can be found in a paper by Chen and Yu [9] (see also a recent paper by Da Lozzo et al. [10] for an application of this decomposition to cubic strong circuit graphs).
Consider a strong circuit graph (G, u, v) such that G is neither a single edge nor a simple cycle. The decomposition distinguishes the case in which the path τ uv (G) coincides with the edge uv (Case A) from the case in which it does not (Case B).
Lemma 5. Suppose that we are in Case A (refer to Fig. 3). Then the graph G = G − uv consists of a sequence of graphs G 1 , . . . , G k , with k ≥ 1, such that: 5a: for i = 1, . . . , k − 1, the graphs G i and G i+1 share a single vertex u i ; further, G i is in the outer face of G i+1 and vice versa in the plane embedding of G; 5b: for 1 ≤ i, j ≤ k with j ≥ i + 2, the graphs G i and G j do not share any vertex; and 5c: for i = 1, . . . , k with u 0 = u and u k = v, (G i , u i−1 , u i ) is a strong circuit graph.
Proof. Consider the BC-tree T of G , which is the tree that is defined as follows. The tree T contains a B-node for each 2-connected component of G and a C-node for each 1-cut of G ; further, T contains an edge between a B-node b and a C-node c if the 1-cut corresponding to c is a vertex of the 2-connected component corresponding to b. First, we have that T is a path. Namely suppose, for a contradiction, that T has a node t with degree at least 3. If t corresponds to a 1-cut {w} of G , as in Fig. 4a, then w belongs to at least three 2-connected components of G and the graph G = G − {w} consists of at least three connected components. Hence, the graph G plus edge uv is disconnected, which implies that {w} is a 1-cut of G; this contradicts Property (a) of (G, u, v). Analogously, if t corresponds to a 2-connected component B of G , as in Fig. 4b, then B contains three distinct 1-cuts {w 1 }, {w 2 }, and {w 3 } of G ; for each i ∈ {1, 2, 3}, the removal of w i from G disconnects G into at least two connected components, at least one of which, denoted by G i , does not contain vertices of B. Since G 1 , G 2 , and G 3 share no vertex, the edge uv connects at most two components G i and G j with i, j ∈ {1, 2, 3}, which implies that {w h } is a 1-cut of G, where h = i, j and h ∈ {1, 2, 3}; this contradicts Property (a) of (G, u, v).
Let G i be the 2-connected component of G corresponding to the B-node b i and let {u i } be the 1-cut of G corresponding to the C-node c i . Then, for i = 1, . . . , k − 1, the graphs G i and G i+1 share a single vertex u i , while for 1 ≤ i, j ≤ k with j ≥ i + 2 the graphs G i and G j do not share any vertex. The vertices u and v are one in G 1 and one in G k ; indeed, if say G 1 did not contain any of u and v, as in Fig. 4c, then {u 1 } would be a 1-cut of G; this would contradict Property (a) of (G, u, v). Assume, w.l.o.g. up to renaming, that u belongs to G 1 and v belongs to G k . We also have that u = u 1 , as if u = u 1 then {u 1 } would be a 1-cut of G, again contradicting Property (a) of (G, u, v); analogously, v = u k−1 .
We prove that G i+1 lies in the outer face of G i in the plane embedding of G, for every i = 1, . . . , k − 1. Suppose for a contradiction that, for some i ∈ {1, . . . , k − 1}, the graph G i+1 lies inside an internal face f of G i (except for the vertex u i , which is on the boundary of f ) in the plane embedding of G. Since the graphs G i+2 , . . . , G k do not share any vertex with G i , by planarity they all lie inside f . It follows that the vertex v lies inside f (note that v = u i even if k = i + 1) and hence it is not incident to the outer face of G, which contradicts Property (b) of (G, u, v). An analogous proof shows that G i lies in the outer face of G i+1 in the plane embedding of G, for every i = 1, . . . , k − 1.
It remains to prove that, for i = 1, . . . , k, the triple (G i , u i−1 , u i ) is a strong circuit graph, where u 0 = u and u k = v. We are going to use the fact that β uv (G) is composed of the paths β uu1 (G 1 ), β u1u2 (G 2 ), . . . , β u k−1 v (G k ). This is because uv coincides with τ uv (G) by Property (c) of (G, u, v) and because G i+1 lies in the outer face of G i and vice versa in the plane embedding of G.
(a) Graph G i is 2-connected by assumption and it is associated with a plane embedding, given that it is a subgraph of the plane graph G.
(b) For i = 1, . . . , k − 1, the vertex u i is external in the plane embedding of G i , since G i is in the outer face of G i+1 and vice versa; analogously, for i = 2, . . . , k, the vertex u i−1 is external in the plane embedding of G i . Further, u 0 = u and u k = v are external in the plane embeddings of G 1 and G k , respectively, since they are external in the plane embedding of G. Finally, for i = 1, . . . , k, the vertices u i−1 and u i are distinct, as otherwise {u i−1 = u i } would be a 1-cut of G, which would contradict Property (a) of (G, u, v).
(c) Suppose, for a contradiction, that the edge u i−1 u i exists and that it does not coincide with τ ui−1ui (G i ). This implies that G i contains vertices different from u i−1 and u i , and hence that no vertex of H i other than u i−1 and u i is incident to the outer face of G, given that all the vertices of H i other than u i−1 and u i lie inside the region delimited by the cycle However, this contradicts Property (d) of (G, u, v).
(d) Consider any 2-cut {a, b} of G i ; then G i has at least two non-trivial {a, b}-components. We prove that a and b are external vertices of G i . Suppose, for a contradiction, that a is an internal vertex of G i (the argument if b is an internal vertex of G i is analogous), as in Proof. We first prove that |V (H)| ≥ 3. Suppose, for a contradiction, that H is a single edge uy 1 . If the degree of u in G is one, then {y 1 } is a 1-cut of G; this contradicts Property (a) for (G, u, v). Otherwise, there is a (H ∪ {v})-bridge B i of G whose attachment in H is u. If B i is trivial, then it coincides with the edge uv; however, this contradicts the hypothesis of Case B. Otherwise, B i is non-trivial; however, this implies that {u, v} is a 2-cut of G, as the removal of u and v from G disconnects y 1 from the vertices in V (B i ) − {u, v}; since neither u nor v is an internal vertex of β uv (G), this contradicts Property (d) for (G, u, v). We now prove the properties of the lemma. First, if G had no H -bridge, then it would not be connected, while it is 2-connected. Hence, G contains distinct H -bridges B 1 , . . . , B with ≥ 1. Each H -bridge B i has at most one attachment y i ∈ V (H), as if B i had at least two attachments in V (H) then it would contain a path (not passing through v) between two vertices of H; however, such a path would be in H, and not in B i , given that H is a maximal 2-connected subgraph of G − {v}. It follows that ≥ 2, as if = 1 then y 1 would be a 1-cut of G, whereas G is 2-connected. Further, for i = 1, 2, . . . , , the vertex v is an attachment of B i , as otherwise y i would be a 1-cut of G, whereas G is 2-connected. Analogously, for i = 1, 2, . . . , , the vertex y i is an attachment of B i , as otherwise v would be a 1-cut of G, whereas G is 2-connected. This proves Property 6a.
Suppose, for a contradiction, that y i = u, for some i ∈ {1, 2, . . . , }. If B i is a trivial Hbridge, then it coincides with the edge uv; however, this contradicts the fact that we are in Case B. If B i is a non-trivial H -bridge, then {u, v} is a 2-cut of G; namely, the removal of u and v from G disconnects the vertices in V (H)−{u} from the vertices in V (B i )−{u, v} -the latter set is non-empty given that B i is non-trivial. However, this contradicts Property (d) for (G, u, v), given that neither u nor v is an internal vertex of β uv (G). It follows that y i = u, for i = 1, 2, . . . , .
We now prove Properties 6b-6f. Since v is incident to the outer face of G, it lies in the outer face of H. It follows that all the H -bridges B 1 , . . . , B lie in the outer face of H, except at the vertices y 1 , . . . , y , respectively. By the planarity of G, there are at most two H -bridges among B 1 , . . . , B that contain edges incident to the outer face of G. If there were only one H -bridge B i containing edges incident to the outer face of G, as in Fig. 7a, then {y i } would be a 1-cut of G, whereas G is 2-connected. Hence, there are exactly two H -bridges among B 1 , . . . , B containing edges incident to the outer face of G. Denote them by B 1 and B , as in Fig. 7b, so that u, y 1 , and y appear in this clockwise order along the outer face of H. Then y 1 = y , as otherwise {y 1 } would be a 1-cut of G, whereas G is 2-connected; in particular, y 1 = y 2 if = 2. It also follows that y 1 is an internal vertex of τ uv (G), that y is an internal vertex of β uv (G) and β uy1 (H), that B 1 contains τ y1v (G), that B contains β y v (G), and that every vertex y i = y 1 , y is not incident to the outer face of G. Now consider any H -bridge B i of G with y i = y . The graph B i is a {y i , v}-component of G, however {y i , v} is a pair of vertices none of which is internal to β uv (G), hence if B i were non-trivial, then Property (d) of (G, u, v) would be violated. It follows that the only H -bridges which might be non-trivial are those whose attachment in H is y ; in particular, since y 1 = y , we have that B 1 is trivial and coincides with the edge y 1 v = τ y1v (G). If there were at least two non-trivial H -bridges whose attachment in H is y , then at least one of them (in fact all the ones different from B ) would not contain any vertex incident to the outer face of G other than y and v; however, this would violate Property (d) of (G, u, v). We now prove that the triple (H, u, y 1 ) is a strong circuit graph. (a) Graph H is 2-connected by assumption and it is associated with a plane embedding, given that it is a subgraph of the plane graph G.
(b) The vertex u is incident to the outer face of H since (G, u, v) satisfies Property (b). The vertex y 1 is a vertex of τ uv (G), as argued above, and hence it is incident to the outer face of G and to the one of H. Finally, u and y 1 are distinct, as otherwise τ uv (G) would coincide with the edge uv, contradicting the fact that we are in Case B.
(c) Suppose, for a contradiction, that the edge uy 1 exists and does not coincide with τ uy1 (H). Then {u, y 1 } is a 2-cut of G, since the removal of u and y 1 disconnects the internal vertices of τ uy1 (H) (which exist since τ uy1 (H) is not the edge uy 1 ) from v. However, none of u and y 1 is an internal vertex of β uv (G); this contradicts Property (d) for (G, u, v).
(d) The proof that (H, u, y 1 ) satisfies Property (d) is very similar to the proof that (G i , u i−1 , u i ) satisfies Property (d) in Lemma 5, hence it is only sketched here. In order to prove Property 6h, assume that B does not coincide with the edge y v, as otherwise there is nothing to prove. Let B be the plane graph obtained by adding the edge y v to B , so that y immediately precedes v in the clockwise order of the vertices along the outer face of B (both y and v are indeed incident to the outer face of B ); see Fig. 8. We prove that (B , y , v) is a strong circuit graph; then Property 6h follows by applying Lemma 5 to (B , y , v).
(a) The graph B is associated with a plane embedding, given that it is a subgraph of the plane graph G. Further, y and v are both incident to the outer face of B , hence the plane graph B is well-defined. We prove that B is 2-connected. Since y and v are adjacent in B , they belong to the same 2-connected component of B . However, the only vertices of B that are incident to edges in E(G) − E(B ) are y and v. It follows that any 1-cut of B is also a 1-cut of G. Then B is 2-connected since G is. This concludes the proof that (B , y , v) is a strong circuit graph, hence it implies Property 6h via Lemma 5. The lemma follows.
We prove that any strong circuit graph (G, u, v) has a planar greedy drawing by exploiting Lemmata 5 and 6 in a natural way. Indeed, if we are in Case A (in Case B) then Lemma 5 (resp. Lemma 6) is applied in order to construct strong circuit graphs for which planar greedy drawings are inductively constructed and then combined together in order to get a planar greedy drawing of (G, u, v). The base cases of the induction are the ones in which G is an edge or a simple cycle. Then a planar greedy drawing of G is directly constructed.
In order to be able to combine planar greedy drawings for the strong circuit graphs (G i , u i−1 , u i ) (and (H, u, y 1 ) if we are in Case B) to construct a planar greedy drawing of (G, u, v), we need the inductively constructed drawings to satisfy some restrictive geometric requirements, which are expressed in the following theorem, which is the core of the proof of Theorem 1.

for every vertex
Before proceeding with the proof of Theorem 7, we comment on its statement. First, let us set δ = 0 and argue about Γ 0 = Γ. Properties 1 and 6 are those that one would expect, as they state that Γ is planar and greedy, respectively. Properties 2 and 3 state that all the edges incident to the outer face of Γ are "close" to horizontal; indeed, the edges of τ uv (G) are horizontal, the edge b 1 b 2 has a slightly negative slope, and all the other edges of β uv (G) have a slightly positive slope. Since Γ is planar, this implies that Γ is contained in a wedge delimited by two half-lines with slopes 0 and −α starting at u. Properties 4 and 5 argue about the existence of certain paths from any vertex to u and v; these two vertices play an important role in the structural decomposition we employ, since distinct subgraphs are joined on those vertices, and the paths incident to them are inductively combined together in order to construct distance-decreasing paths. Finally, all these properties still hold true if u is moved by an arbitrary non-negative amount δ to the left. This is an important feature we exploit in one of our inductive cases.
We now present an inductive proof of Theorem 7. In the base cases G is a single edge (we call this the Trivial Case) or a simple cycle (we call this the Cycle Case).
We start with the Trivial Case, in which G is a single edge. Although Theorem 7 assumes that |V (G)| ≥ 3, for its proof we need to inductively draw certain subgraphs of G which might be single edges. Whenever we need to draw a strong circuit graph (G, u, v) such that G is a single edge uv, we draw it as a horizontal straight-line segment with positive length, with u to the left of v. We remark that, since Theorem 7 assumes that |V (G)| ≥ 3, we do not need the constructed drawing to satisfy Properties 1-6.
We next deal with the Cycle Case, in which G is a simple cycle with at least 3 vertices. Refer to Fig. 10. By Property (d) of (G, u, v), the set {u, v} is not a 2-cut of G, hence u and v appear consecutively along the cycle G. By Property (c) of (G, u, v), the edge uv coincides with the path τ uv (G). Drawing Γ is constructed as follows.
This completes the construction of Γ. We have the following.
. This implies that Γ δ satisfies Property 3. Concerning Property 4, let x = b i with i < m. Then a path P x satisfying the requirements can be defined as P In the former case, the only edge of P x has slope 0 ∈ (−α; α); in the latter case, all the edges of P x have slope α 2 ∈ (−α; α) and P x does not pass through u. Hence Γ δ satisfies Property 4. Concerning Property 5, let x = b i with i > 1. Then a path Q x satisfying the requirements can be defined as δ+d H (Γ,b1b2) , which is smaller than π, given that . This implies that Γ δ satisfies Property 5. Finally we deal with Property 6. Let x = b i and y = b j , for some 1 ≤ i, j ≤ m.
is distance-decreasing can be proved as in the previous point. If j = 1 and i ≥ 3, then the path In order to prove that, it suffices to argue that d( This concludes the proof of the lemma. We now discuss the inductive cases. In Case A the path τ uv (G) coincides with the edge uv, while in Case B it does not. We discuss Case A first. Let G = G − uv, where G consists of a sequence of graphs G 1 , . . . , G k , with k ≥ 1, satisfying the properties described in Lemma 5. Our construction is different if k = 1 and k ≥ 2.
Suppose first that k = 1; by Lemma 5 the triple (G = G 1 , u, v) is a strong circuit graph (and G 1 is not a single edge, as otherwise we would be in the Trivial Case). Apply induction in order to construct a straight-line drawing Γ of G with α 2 as a parameter. Let τ uv (G ) = (u = a 1 , a 2 , . . . , a t = v). By Property 2 the path τ uv (G ) lies on a horizontal line u in Γ with u to the left of v. Let Y > 0 be the minimum distance in Γ of any vertex strictly below u from u . Let We construct a straight-line drawing Γ of G from Γ as follows; refer to Fig. 11. Decrease the y-coordinate of the vertex a 2 by ε. Further, decrease the y-coordinate of the vertex a i , with i = 3, 4, . . . , t − 1, so that it ends up on the straight-line segment a 2 a t . Draw uv as a straight-line segment. We have the following.
d(Γ , a 2 a t ) Figure 11 The straight-line drawing Γ of G in Case A if k = 1.
Proof. Concerning Property 1, note first that Γ is planar, given that ε < ε * Γ . Since Γ δ and Γ coincide, except for the position of the vertex u, we only need to prove that no edge incident to u crosses any other edge in Γ δ . Then consider any two edges uu and ww with u , w, w ∈ V (G) (possibly w = u or w = u ) and suppose, for a contradiction, that they cross or overlap in Γ δ . u u u δ u w u ε u ε w Figure 12 Illustration for the proof that the edges uu and ww do not cross in Γ δ . Fig. 12. If u = v, then uu and ww do not cross in Γ δ , given that no vertex other than u and v lies on or above u in Γ δ . We can hence assume that u = v and that y(u ) < y(u). By Properties 1-3, we have that Γ lies in the closed wedge that is delimited by the half-lines starting at u with slopes 0 and − π 4 . It follows that x(u ) > x(u) in Γ , Γ, and Γ δ (given that every vertex has the same x-coordinate in Γ , Γ, and Γ δ , except for u, whose x-coordinate might be smaller in Γ δ than in Γ and Γ). Consider the unbounded region R of the plane that is delimited by u from above, by the horizontal line u through u from below, and by the representation of the edge uu in Γ from the right. For any value δ > 0, we have that uu lies in the interior of R (except at points u and u ) in Γ δ , hence if uu and ww cross in Γ δ then at least one end-vertex of ww , say w, lies in the interior of R, given that y(w), y(w ) ≤ y(u) and that ww does not cross uu in Γ. This implies that x(u) < x(w) < x(u ) in Γ , Γ, and Γ δ . We now distinguish four cases, based on whether u and/or w belong to V (τ uv (G )).

Refer to
If u , w ∈ V (τ uv (G )), then by Property 2 we have that u and w lie on u in Γ . However, since x(u) < x(w) < x(u ), it follows that the edge uu overlaps the vertex w in Γ , a contradiction to Property 1 of Γ . If u ∈ V (τ uv (G )) and w / ∈ V (τ uv (G )), then when transforming Γ into Γ the ycoordinate of u has been decreased by a value ε u ≤ ε which is larger than the distance ε w ≥ Y between w and u . This contradicts ε ≤ Y 2 < Y . If u / ∈ V (τ uv (G )) and w ∈ V (τ uv (G )), then when transforming Γ into Γ the ycoordinate of w has been decreased by a value ε w ≤ ε. The point p on the edge uu with x-coordinate equal to x(w) has y-coordinate larger than y(w), hence the distance from p to u is a value ε p < ε w . This implies that the drawing obtained from Γ by decreasing the y-coordinate of w by ε p , while every other vertex stays put, is not planar, given that the edge uu overlaps the vertex w. However, since ε p < ε w , this contradicts ε w ≤ ε < ε * Γ . Finally, if u , w / ∈ V (τ uv (G )), then u, u and w have the same positions in Γ and Γ. Consider the line through u and w; let q be its intersection point with u and let δ q be the Euclidean distance between q and u in Γ . Then the drawing Γ δq is not planar as the edge uu overlaps the vertex w. This contradicts Property 1 of Γ .
Concerning Property 2, note that u and v lie on the same horizontal line u (with u to the left of v) in Γ since they do in Γ and since they have not been moved when transforming Γ into Γ. Since τ uv (G) coincides with the edge uv, it follows that Γ δ satisfies Property 2.
Property 3 is satisfied by Γ δ since it is satisfied by Γ δ and since no vertex of β uv (G ) = β uv (G) moves when transforming Γ into Γ (indeed, τ uv (G ) and β uv (G ) do not share any vertex other than u and v, given that G is 2-connected).
We now discuss Property 4. Let x ∈ V (G). If x = u, let P x = (u, v); then the only edge of P x has slope 0 ∈ (−α; α). If x = u, then let P , for i = 1, . . . , p − 1, and such that u / ∈ V (P x ). This path exists since Γ satisfies Property 4, by induction. We distinguish two cases.
If no vertex of P x −{v} belongs to τ uv (G ), then P x = P x satisfies the required properties. Indeed, no vertex other than those internal to τ uv (G ) moves when transforming Γ into Γ and no vertex other than u moves when transforming Γ into Γ δ ; thus, P x has the same representation (and in particular each edge of P x has the same slope) in Γ and Γ δ . Otherwise, a vertex of P x − {v} belongs to τ uv (G ); let h be the smallest index such that v h = a j , for some a j ∈ V (τ uv (G )) − {v} and define P x = (x = v 1 , v 2 , . . . , v h = a j , a j+1 , . . . , a t = v). Refer to Fig. 13. Note that u / ∈ V (P x ), given that u / ∈ V (P x ). Hence, it suffices to argue about the slopes of the edges of P x in Γ (rather than in Γ δ ). For i = 1, . . . , h − 2, the slope of the edge v i v i+1 is in (−α; α) in Γ since it is in (− α 2 ; α 2 ) ⊂ (−α; α) in Γ and since neither v i nor v i+1 moves when transforming Γ into Γ. Further, for i = j, . . . , t − 1, the slope of the edge a i a i+1 in Γ is arctan ε d(Γ ,a2at) , which is in the interval (0; α) ⊂ (−α; α), given that ε, d(Γ , a 2 a t ) > 0 and that ε < tan(α) · d(Γ , a 2 a t ). Finally, let s and s be the slopes of the edge v h−1 v h in Γ and Γ, respectively.
Finally, we deal with Property 6. Consider any two vertices x, y ∈ V (G).
First, assume that x, y = u. By induction, there exists a path P xy from x to y in G that is distance-decreasing in Γ with u / ∈ V (P xy ). By Lemma 2 and since, for every vertex z ∈ V (G), the Euclidean distance between the positions of z in Γ and Γ is at most ε < ε * Γ , we have that P xy is also distance-decreasing in Γ. Further, since all the vertices other than u have the same position in Γ and Γ δ , it follows that P xy is a distance-decreasing path from x to y not passing through u in Γ δ . Second, suppose that y = u. Consider the path Q x in G from Property 5, whose every edge has slope in (π − α; π + α) in Γ δ . Since α ≤ π 4 , it follows that Q x is a π-path (according to the definition in [11]) or is π-monotone (according to the definition in [7]), where for some angle β a path (q 1 , q 2 , . . . , q r ) is a β-path or equivalently is β-monotone if every edge q i q i+1 has slope in the interval (β − π 4 ; β + π 4 ). In [11,Lemma 3] it is proved that a β-path is distance-decreasing (in fact, it satisfies a much stronger property, namely it is increasing-chord); hence, Q x is distance-decreasing in Γ δ . Finally, suppose that x = u and consider a path P xy from x to y in G that is distancedecreasing in Γ . We prove that P xy is distance-decreasing in Γ δ , as well. Let xx be the edge of P xy incident to x. Differently from the case in which x, y = u, we cannot directly apply Lemma 2, given that it is not guaranteed that ε < ε * Γ δ . However, since for every vertex z ∈ V (P xy ) the Euclidean distance between the positions of z in Γ and Γ is at most ε < ε * Γ , by Lemma 2 we have that P xy is distance-decreasing in Γ. Further, the path obtained from P xy by removing the vertex x = u and the edge xx has the same representation in Γ δ and Γ, given that it does not contain u, hence it is distance-decreasing in Γ δ . Thus, it only remains to show that d(Γ δ , xy) > d(Γ δ , x y). First, since x = u, we have that x , y = u. Hence, d(Γ δ , x y) = d(Γ, x y). Second, denote by u Γ and u Γ δ the positions of u in Γ and Γ δ , respectively. By Properties 1-3, the entire drawing Γ, and in particular vertex y, lies in the closed wedge that is delimited by the half-lines starting at u Γ and with slopes 0 and −α. Then the angle incident to u Γ in the triangle ∆yu Γ u Γ δ is at least π − α > π 2 , hence the straight-line segment between u Γ δ and y is the longest side of that triangle. It follows that d(Γ δ , xy) ≥ d(Γ, xy). Thus, x y), where the second inequality holds true since P xy is distance-decreasing in Γ. Hence, P xy is distance-decreasing in Γ δ . This concludes the proof of the lemma.
We now discuss the case in which k ≥ 2. Refer to Fig. 15. By Lemma 5, for i = 1, . . . , k, the triple (G i , u i−1 , u i ) is a strong circuit graph, where u 0 = u, u k = v, and u i is the only vertex shared by G i and G i+1 , for i = 1, . . . , k − 1. Figure 15 The straight-line drawing Γ of G in Case A if k ≥ 2. In this example k = 4. The gray angle in the drawing is α 2 .
If G 1 is a single edge, then apply induction in order to construct a straight-line drawing Γ 1 of G 1 and define ε = 1 2 min{ε * Γ1 , tan(α) · d(Γ 1 , u 0 u 1 )}. If G 1 is not a single edge, then apply induction in order to construct a straight-line drawing Γ 1 of G 1 with α 2 as a parameter. By Property 2 of Γ 1 , the path τ u0u1 (G 1 ) lies on a horizontal line u . Let Y > 0 be the minimum distance in Γ 1 of any vertex strictly below u from u . Let ε = 1 2 min{ε * Γ1 , Y, tan(α) · d(Γ 1 , u 0 u 1 )}. In both cases, decrease the y-coordinate of u 1 by ε. Further, decrease the y-coordinate of every internal vertex of the path τ u0u1 (G 1 ), if any, so that it ends up on the straight-line segment u 0 u 1 . Now consider a half-line h with slope s = α 2 starting at u 1 . Denote by p v the point at which h intersects the horizontal line u through u. For i = 2, . . . , k, apply induction in order to construct a straight-line drawing Γ i of G i with α 3 as a parameter (if G i is a single edge, then the parameter does not matter). Uniformly scale the drawings Γ 2 , . . . , Γ k so that the Euclidean distance between u i−1 and u i is equal to d(Γ1,u1pv) k−1 . For i = 2, . . . , k, rotate the scaled drawing Γ i around u i−1 counter-clockwise by s radians. Translate the scaled and rotated drawings Γ 2 , . . . , Γ k so that the representations of u i in Γ i and Γ i+1 coincide, for i = 1, . . . , k − 1. Finally, draw the edge uv as a straight-line segment. This completes the construction of a drawing Γ of G. We have the following. Proof. Throughout the proof, we denote by Γ 1,δ the drawing obtained from Γ 1 by moving the position of the vertex u 0 = u by δ units to the left (where Γ 1 is understood as the drawing of G 1 in which the vertices of τ u0u1 (G 1 ) all lie on u ).
We first prove Property 2. Because of the uniform scaling which has been applied to Γ 2 , . . . , Γ k , we have that d(Γ, u i−1 u i ) = d(Γ,u1pv) k−1 for i = 2, . . . , k. Since the vertices , which implies that u and v lie on u in Γ and Γ δ . Further, u is to the left of v in Γ, given that Γ 1 satisfies Property 2 and that 0 < s < π 2 . Since τ uv (G) coincides with the edge uv, it follows that Γ δ satisfies Property 2.
We next prove that Γ δ satisfies Property 3. Observe that β uv We first argue about the slope of the edge b 1 b 2 .
3 ) by Property 3 of Γ i , plus s, which results from the rotation of Γ i . Hence s j ∈ ( α 2 ; 5α 6 ) ⊂ (0; α). If b j b j+1 belongs to a graph G i with |V (G i )| ≥ 3, with i ≥ 2, and with b j = u i−1 , then s j is given by the slope b j b j+1 has in Γ i , which is in (− α 3 ; 0) by Property 3 of Γ i , plus s, which results from the rotation of Γ i . Hence s j ∈ ( α 6 ; α 2 ) ⊂ (0; α). If b j b j+1 belongs to G 1 , if |V (G 1 )| ≥ 3, and if b j+1 = u 1 , then since Γ 1,δ satisfies Property 3 and since u 1 is the only vertex of β u0u1 (G 1 ) whose positions in Γ 1,δ and Γ δ do not coincide, it follows that s j ∈ (0; α) since the slope of b j b j+1 in Γ 1,δ is in (0; α). Finally, assume that b j b j+1 belongs to G 1 , that |V (G 1 )| ≥ 3, and that b j+1 = u 1 . Note that b j = u, given that j ≥ 2, hence by Property 3 of Γ 1 we have that x(b j ) < x(b j+1 ) and that y(b j ) < y(b j+1 ) in Γ 1 . Note that the positions of b j in Γ 1 and Γ δ coincide, given that b j / ∈ V (τ u0u1 (G 1 )); further, b j+1 moves down by ε when transforming Γ 1 into Γ δ , however its x-coordinate stays unchanged; this implies that s j is smaller than the We now prove Property 1. First, the edge uv does not cross or overlap any other edge of G, since no vertex other than u and v lies on or above u in Γ and Γ δ . Hence, we only need to argue about crossings among edges in the graphs G 1 , . . . , G k .
We first deal with Γ. For i = 1, . . . , k, the inductively constructed drawing Γ i of G i is planar, by Property 1. Further, for i = 2, . . . , k, the drawing of G i in Γ is congruent to Γ i , up to affine transformations (a uniform scaling, a rotation, and a translation), which preserve planarity. Moreover, since ε < ε * Γ1 , by Lemma 2 we have that the drawing of G 1 in Γ is planar, as well. It follows that no two edges in the same graph G i cross each other in Γ, for each i = 1, . . . , k. Since Γ satisfies Property 3, the path β uv (G) is represented in Γ by a curve monotonically increasing in the x-direction from u to v. Further, the path τ = k i=1 τ ui−1ui (G i ) is also represented in Γ by a curve monotonically increasing in the x-direction from u to v, since it is composed of the straight-line segment u 0 u 1 , which has slope − arctan ε d(Γ1,u0u1) ∈ (− π 2 ; 0), and of the straight-line segment u 1 u k , which has slope s = α 2 ∈ (0; π 2 ). Hence, for i = 1, 2, . . . , k − 1, the vertical line through u i has the drawings of G 1 , . . . , G i to its left and the drawings of G i+1 , . . . , G k to its right in Γ. It follows that no two edges in distinct graphs G i and G j cross in Γ. This proves the planarity of Γ.
Since Γ δ and Γ coincide, except for the position of u, it remains to prove that no edge uu incident to u with u = v crosses or overlaps any other edge in Γ δ . Since the vertical line through u 1 has the drawing of G 1 to its left and the drawings of G 2 , . . . , G k to its right in Γ δ , such a crossing might only occur between uu and another edge ww of G 1 . The proof that uu and ww do not cross or overlap is the same as in the proof of Lemma 9, with G 1 playing the role of G and Γ 1 playing the role of Γ .
in Γ 1 and Γ (note that the x-coordinates of the vertices do not change when transforming Γ 1 into Γ). Further, by Properties 1-4 of Γ 1 , we have that v h−1 lies below u , which contains v h ; hence, u); then the only edge of Q x has slope π ∈ (π − α; π + α) in Γ δ . If x = u i , for some i ∈ {1, . . . , k − 1}, then let Q x = i j=1 β uj uj−1 (G j ); recall that β uj uj−1 (G j ) has the same vertices as τ uj−1uj (G j ), however in the reverse linear order. Denote the vertices of Q x by ( x = w 1 , w 2 , . . . , w q = u). Consider the edge w l w l+1 , for any 1 ≤ l ≤ q − 1.
Assume first that i ≥ 2. Since Γ i satisfies Property 5, there exists a path Q i x from x to u i−1 in G i whose every edge has slope in (π − α 3 ; π + α 3 ) in Γ i ; then Q x consists of Q i x and of the path i−1 j=1 β uj uj−1 (G j ); denote the vertices of Q x by (x = w 1 , w 2 , . . . , w q = u). Consider the edge w l w l+1 , for any 1 ≤ l ≤ q − 1.
If w l w l+1 is in G 1 and l ≤ q − 2, then its slope in Γ δ is π − arctan ε d (Γ1,u0u1) , which is in the interval (π − α; π + α), as proved in the case x = u i .
We finally deal with Property 6. Consider any two distinct vertices x, y ∈ V (G). If x and y belong to the same graph G i , for some i ∈ {1, . . . , k}, then there exists a distance-decreasing path P xy from x to y in Γ i , given that Γ i satisfies Property 6. If i ∈ {2, . . . , k}, the drawing of G i in Γ δ is congruent to Γ i , up to three affine transformations (a uniform scaling, a rotation, and a translation) that preserve the property of a path to be distance-decreasing; hence P xy is distance-decreasing in Γ δ as well. If i = 1, then the proof that P xy is distance-decreasing in Γ δ is the same as the proof that Γ δ satisfies Property 6 in Lemma 9, with Γ 1 playing the role of Γ .
We can hence assume that x and y belong to two distinct graphs G i and G j , respectively.
Suppose first that 2 ≤ i < j ≤ k. Then let P xy be the path composed of: a path P i x in G i from x to u i whose every edge has slope in (− α 3 ; α 3 ) in Γ i ; the path j−1 l=i+1 τ u l−1 u l (G l ); and a path P uj−1y in G j that is distance-decreasing in Γ j . By induction, the paths P i x and P uj−1y exist since Γ i satisfies Property 4 and Γ j satisfies Property 6, respectively. We prove that P xy is distance-decreasing in Γ δ ; note that u / ∈ V (P xy ). Let P xy = (z 1 , z 2 , . . . , z s ); then we need to prove that d(Γ δ , z h z s ) > d(Γ δ , z h+1 z s ), for h = 1, 2, . . . , s − 2. We distinguish three cases.
If z h z h+1 is in G j , then (z h , z h+1 , . . . , z s ) is a sub-path of P uj−1y , hence it is distancedecreasing in Γ δ since it is distance-decreasing in Γ j and since the drawing of G j in Γ δ is congruent to Γ j , up to three affine transformations (a uniform scaling, a rotation, and a translation) that preserve the property of a path to be distance-decreasing. If z h z h+1 is in τ u l−1 u l (G l ), for some l ∈ {i+1, i+2, . . . , j −1}, as in Fig. 17, then it has slope s = α 2 . Consider the line h with slope π+α 2 through u l , oriented towards increasing y-coordinates. By Lemma 3, this line has the drawings of G l+1 , G l+2 , . . . , G k to its right; this is because by Property 3 of Γ δ every edge in β u l v (G) has slope in the interval (0; α), where −π+α 2 < 0 < α < π+α 2 , and because the path Further, by Lemma 3, the line h has the drawing of the path β u l u l−1 (G l ) to its left; this is because every edge in β u l u l−1 (G l ) has slope s = π + α 2 , where π+α 2 < π + α 2 < 3π+α 2 . Then the line h parallel to h , passing through the midpoint of the edge z h z h+1 , and oriented towards increasing y-coordinates has h to its right, given that the path β u l u l−1 (G l ) (and in particular the midpoint of the edge z h z h+1 ) is to the left of h , hence h has the drawings of G l+1 , G l+2 , . . . , G k (and in particular the vertex z s ) to its right. Since the half-plane to the right of h represents the locus of the points of the plane that are closer to z h+1 than to z h , it follows that d(Γ δ , z h z s ) > d(Γ δ , z h+1 z s ). If z h z h+1 is in P i x , as in Fig. 18, then by Property 4 it has slope in (− α 3 ; α 3 ) in Γ i . Since Γ i is counter-clockwise rotated by s radians in Γ δ , it follows that z h z h+1 has slope in (s − α 3 ; s + α 3 ) = ( α 6 ; 5α 6 ) in Γ δ . Consider the line h that passes through u i , that is directed towards increasing y-coordinates and that is orthogonal to the line through z h and z h+1 . Denote by s h the slope of h . Then s h ∈ ( π 2 + α 6 ; π 2 + 5α 6 ). By

On Planar Greedy Drawings of 3-Connected Planar Graphs
Lemma 3, the line h has the drawings of G i+1 , . . . , G k to its right; this is because by Property 3 of Γ δ every edge in β uiv (G) has slope in (0; α) with s h −π < − π 2 + 5α 6 < 0 < α < π 2 + α 6 < s h and because the path Further, by Lemma 3, the line h has the drawings of G 2 , . . . , G i to its left; this is because by Property 3 of Γ δ every edge in τ uiu1 (G) has slope in (π; π+α) with s h < π 2 + 5α 6 < π < π+α < 3π 2 + α 6 < π+s h and because the path i m=2 β umum−1 (G m ) has slope s = π+ α 2 , where s h < π 2 + 5α 6 < π+ α 2 < 3π 2 + α 6 < π+s h . Now consider the line h parallel to h , passing through the midpoint of the edge z h z h+1 , and oriented towards increasing y-coordinates. This line has h to its right, given that the drawing of G i (and in particular the midpoint of z h z h+1 ) is to the left of h in Γ δ . Thus, h has the drawings of G i+1 , G i+2 , . . . , G k (and in particular the vertex z s ) to its right. Since the half-plane to the right of h represents the locus of the points of the plane that are closer to z h+1 than to z h , it follows that Figure 18 Illustration for the proof that d( The case in which 2 ≤ j < i ≤ k is symmetric to the case in which 2 ≤ i < j ≤ k. Suppose next that i = 1 and j > 1. Then let P xy be the path composed of: a path P 1 x in G 1 from x to u 1 whose every edge has slope in (−α; α) in Γ δ , where P 1 x does not pass through u, unless x = u; the path j−1 l=2 τ u l−1 u l (G l ); and a path P uj−1y in G j that is distance-decreasing in Γ j . The path P uj−1y exists by induction since Γ j satisfies Property 6. We prove that a path P 1 x satisfying the above properties exists in Γ δ . Let τ u0u1 (G 1 ) = (u 0 = c 1 , c 2 , . . . , c r = u 1 ). If x = u, then let P 1 x = τ u0u1 (G 1 ). Every edge of P 1 x other than c 1 c 2 has slope − arctan ε d (Γ1,u0u1) in Γ δ , while c 1 c 2 has slope − arctan ε δ+d (Γ1,u0u1) . These slopes are smaller than 0, given that ε, d(Γ 1 , u 0 u 1 ) > 0 and δ ≥ 0, and larger than −α, given that δ ≥ 0 and ε < tan α · d(Γ 1 , u 0 u 1 ). Thus, every edge of P 1 x has slope in (−α; α) in Γ δ . If x = u, then P 1 x can be shown to exist as in the proof that Γ δ satisfies Property 4, by considering a path (x = v 1 , v 2 , . . . , v p = u 1 ) in G 1 that does not pass through u and whose every edge has slope in (− α 2 ; α 2 ) in Γ 1 and by defining , c j+1 , . . . , c r ), where h is the smallest index such that v h ∈ V (τ u0u1 (G 1 )). This concludes the proof that a path P 1 x satisfying the required properties exists in Γ δ . Note that u / ∈ V (P xy ), unless x = u, given that u / ∈ V (P 1 x ), unless x = u. Let P xy = (z 1 , z 2 , . . . , z s ); we prove that, for any h = 1, 2, . . . , s − 2, it holds true that d(Γ δ , z h z s ) > d(Γ δ , z h+1 z s ). This can be proved exactly as in the case 2 x and recall that the slope of every edge of P 1 x in Γ δ is in (−α; α). Similarly to the case 2 ≤ i < j ≤ k, consider the line h that passes through u 1 , that is directed towards increasing y-coordinates and that is orthogonal to the line through z h and z h+1 . Denote by s h the slope of h . Then s h ∈ ( π 2 − α; π 2 + α). By Lemma 3, the line h has the drawings of G 2 , . . . , G k to its right; this is because by Property 3 of Γ δ every edge in β u1v (G) has slope in (0; α) with s h − π < − π 2 + α < 0 < α < π 2 − α < s h and because the path Further, by Lemma 3, the line h has the drawing of G 1 to its left; this is because by Property 3 of Γ 1 every edge in τ u1u0 (G) has slope in (π − α 2 ; π + α 2 ), where s h < π 2 + α < π − α 2 < π + α 2 < 3π 2 − α < π + s h and because every edge of the path β u1u0 (G 1 ) has slope either π − arctan ε d (Γ1,u0u1) . Now consider the line h parallel to h , passing through the midpoint of the edge z h z h+1 , and oriented towards increasing y-coordinates. This line has h to its right, given that the drawing of G 1 (and in particular the midpoint of z h z h+1 ) is to the left of h in Γ δ . Thus, h has the drawings of G 2 , G 3 , . . . , G k (and in particular the vertex z s ) to its right. Since the half-plane to the right of h represents the locus of the points of the plane that are closer to z h+1 than to z h , it follows that d(Γ δ , z h z s ) > d(Γ δ , z h+1 z s ). Suppose finally that i > 1 and j = 1. Then P xy consists of three paths, one contained in G i , one coinciding with i−1 l=2 β u l u l−1 (G l ), and one contained in G 1 .
The first path in P xy is a path Q i x in G i from x to u i−1 whose every slope in Γ i is in (π − α 3 ; π + α 3 ); this path exists since Γ i satisfies Property 5. Since Γ i is counterclockwise rotated by s radians in Γ δ , it follows that every edge of Q i x has slope in (π + s − α 3 ; π + s + α 3 ) = (π + α 6 ; π + 5α 6 ) in Γ δ . We prove that, for every edge z h z h+1 of Q i x , it holds true that d(Γ δ , z h y) > d(Γ δ , z h+1 y). Consider the line h that passes through u i−1 , that is directed towards increasing y-coordinates and that is orthogonal to the line through z h and z h+1 . Denote by s h the slope of h . Then s h ∈ ( π 2 + α 6 ; π 2 + 5α 6 ). By Lemma 3, the line h has the drawing of G i to its right; this is because by Property 3 of Γ δ every edge in β ui−1ui (G) has slope in (−α; α) with s h − π < − π 2 + 5α 6 < −α < α < π 2 + α 6 < s h and because the path τ ui−1ui (G i ) has slope s = α 2 , where s h − π < − π 2 + 5α 6 < α 2 < π 2 + α 6 < s h . Further, by Lemma 3, the line h has the drawings of G 1 , . . . , G i−1 (and in particular y) to its left; this is because by Property 3 of Γ δ every edge in τ ui−1u0 (G) has slope in (π − α; π + α) where s h < π 2 + 5α 6 < π − α < π + α < 3π 2 + α 6 < π + s h and because every edge of the path i−1 m=1 β umum−1 (G m ) has slope either s = π + α 2 , or π − arctan ε d (Γ1,u0u1) , or π − arctan ε δ+d(Γ1,u0u1) , where s h < π 2 + 5α 6 < π − α < π − arctan ε d(Γ1,u0u1) ≤ π − arctan ε δ+d(Γ1,u0u1) < π < π + α 2 < 3π 2 + α 6 < π + s h . Now consider the line h parallel to h , passing through the midpoint of the edge z h z h+1 , and oriented towards increasing y-coordinates. This line has h to its left, given that the drawing of G i (and in particular the midpoint of z h z h+1 ) is to the right of h in Γ δ . Thus, h has the drawings of G i−1 , G i−2 , . . . , G 1 (and in particular the vertex y) to its left. Since the half-plane to the left of h represents the locus of the points of the plane that are closer to z h+1 than to z h , it follows that d(Γ δ , z h y) > d(Γ δ , z h+1 y).
The second path in P xy is . Consider an edge z h z h+1 of this path in a graph G l , for some l ∈ {2, . . . , i − 1}. Then z h z h+1 has slope π + s = π + α 2 . Consider the line h that has slope s h = π+α 2 , that passes through u l−1 , and that is oriented towards increasing y-coordinates. By Lemma 3, the line h has the drawing of G l to its right; this is because by Property 3 of Γ δ every edge in β u l−1 u l (G) has slope in (0; α) with s h − π = −π+α 2 < 0 < α < π+α 2 = s h and because the path Further, by Lemma 3, the line h has the drawings of G l−1 , G l−2 , . . . , G 1 to its left; this is because by Property 3 of Γ δ every edge in τ u l−1 u0 (G) has slope in (π − α; π + α), where s h = π+α 2 < π − α < π + α < 3π+α 2 = s h + π and because every edge of the path Now consider the line h parallel to h , passing through the midpoint of the edge z h z h+1 , and oriented towards increasing y-coordinates. This line has h to its left, given that the drawing of G l (and in particular the midpoint of z h z h+1 ) is to the right of h in Γ δ . Thus, h has the drawings of G l−1 , G l−2 , . . . , G 1 (and in particular vertex y) to its left. Since the half-plane to the left of h represents the locus of the points of the plane that are closer to z h+1 than to z h , it follows that d(Γ δ , z h y) > d(Γ δ , z h+1 y). The third path P u1y in P xy is defined as follows. If y = u, let P u1y = β u1u0 (G 1 ). Then every edge of P u1y has slope either π−arctan ε d (Γ1,u0u1) or π−arctan ε δ+d (Γ1,u0u1) . Both these slopes are smaller than π, given that ε, d(Γ 1 , u 0 u 1 ) > 0 and δ ≥ 0, and larger than π − α, given that δ ≥ 0 and ε < tan(α) · d(Γ 1 , u 0 u 1 ). Thus, P u1y is a π-path, and hence it is distance-decreasing (see [11] and the proof of Property 6 in Lemma 9). If y = u, then let P u1y be a distance-decreasing path in Γ 1 not passing through u. This path exists by induction, given that Γ 1 satisfies Property 6. Since P u1y does not pass through u, it has the same representation in Γ δ and Γ. Since the Euclidean distance between the positions of any vertex of G 1 in Γ 1 and Γ is at most ε < ε * Γ1 , by Lemma 2 we have that P u1y is distance-decreasing in Γ and hence in Γ δ .
Hence Γ δ satisfies Property 6. This concludes the proof of the lemma.
We now discuss Case B, in which (G, u, v) is decomposed according to Lemma 6. Refer  Figure 19 The straight-line drawing Γ of G in Case B. For the sake of readability, φ and ρ are larger than they should be. The dark gray angle is equal to β.
. Note that β > 0, given that We claim that v lies to the right of y 1 . The polygonal line representing β y y1 (H) in Γ H and the straight-line segment y v are both incident to y . By definition of φ and since Γ H satisfies Property 3, β y y1 (H) is composed of straight-line segments with slopes in the range [φ; α 2 ), while y v has slope β. The claim then follows from 0 < β < φ < π 2 . Denote by d y1v the distance between y 1 and v. Let Y > 0 be the minimum distance in Γ H of any vertex strictly below u from u .
Let ρ = min{ Let D ρ be the disk with radius ρ centered at v. Let p ρ,β (p ρ,u ) be the intersection point closer to y (resp. to y 1 ) of the boundary of D ρ with h β (resp. with u ). Let d * be the Euclidean distance between y and p ρ,β .

Figure 21
The drawing Γj and the disk Dj centered at uj with radius d (Γj, uj−1uj).
; hence β uj−1uj (G j ) also lies inside D j . By Property 1 of Γ j , the entire drawing Γ j lies inside D j . Hence, u j−1 is the farthest vertex of G j from u j in Γ j . This property holds true also after the drawings Γ 1 , . . . , Γ k are uniformly scaled; further, after the scaling, the distance between u j−1 and u j is ρ k , by construction. By the triangular inequality, we have We now discuss the possible crossings that might occur in Γ δ .
The drawing of H in Γ δ coincides with Γ H,δ , hence it is planar since Γ H,δ satisfies Property 1 by induction. Analogously, the drawings of G 1 , G 2 , . . . , G k in Γ δ are planar since they coincide with Γ 1,d * , Γ 2 , . . . , Γ k , which satisfy Property 1 by induction. Since Γ δ satisfies Property 3, the path β y v (G) is represented in Γ δ by a curve monotonically increasing in the x-direction from y to v. Further, the path τ = k i=1 τ ui−1ui (G i ) is represented in Γ δ by a straight-line segment with slope β ∈ (0; α 4 ) ⊂ (0; π 16 ). Hence, for i = 1, . . . , k − 1, the vertical line through u i has the drawings of G 1 , . . . , G i to its left and those of G i+1 , . . . , G k to its right in Γ δ . It follows that no two edges in distinct graphs G i and G j cross in Γ δ . Consider the vertical line 1 through y 1 . By Properties 1-3 of Γ H,δ , the line 1 has Γ H,δ to its left. Further, since ρ < d y1v , the disk D ρ lies to the right of 1 . Since all the vertices different from u 0 of the graphs G 1 , . . . , G k lie inside D ρ , it follows that no edge in a graph G 1 , . . . , G k crosses an edge of H, unless the former is incident to u 0 . However, all the edges in G 1 , . . . , G k that are incident to u 0 (in fact only G 1 contains such edges) have slope at most β, as they lie on or below h β . Hence they all lie to the right of the path β b2bm (H) and do not cross edges of H in Γ δ .
Finally, we deal with Property 6. Consider any two vertices x, y ∈ V (G). We prove the existence of a path P xy from x to y in G which does not pass through u, unless x = u or y = u, and which is distance-decreasing in Γ δ . We distinguish several cases, based on which graphs among H, G 1 , . . . , G k the vertices x and y belong to.
Suppose first that x and y belong to H. Since Γ H,δ satisfies Property 6, there exists a path P xy from x to y in H which does not pass through u, unless x = u or y = u, and which is distance-decreasing in Γ H,δ . Since the drawing of H in Γ δ coincides with Γ H,δ , it follows that P xy is distance-decreasing in Γ δ . Suppose next that x and y belong to the same graph G i , for some i ∈ {1, 2, . . . , k}. Since the drawing Γ i (or Γ 1,d * if i = 1) satisfies Property 6, there exists a path P xy from x to y in G i that is distance-decreasing in Γ i (in Γ 1,d * if i = 1). Since the drawing of G i in Γ δ is congruent to Γ i (to Γ 1,d * if i = 1) up to three affine transformations, namely a uniform scaling, a rotation, and a translation, that preserve the property of a path to be distance-decreasing, it follows that P xy is distance-decreasing in Γ δ . Note that u / ∈ V (P xy ). Suppose now that x belongs to a graph G i and y belongs to a graph G j for some 1 ≤ i < j ≤ k. Then let P xy be the path composed of a path P x in G i from x to u i whose every slope in Γ i is in (−α ; α ), of the path j−1 l=i+1 τ u l−1 u l (G l ), and of a path P uj−1y in G j that is distance-decreasing in Γ j . The path P x exists since Γ i (Γ 1,d * if i = 1) satisfies Property 4; the path P uj−1y exists since Γ j satisfies Property 6. The proof that P xy is distance-decreasing in Γ δ is the same as the one that P xy is distancedecreasing in Γ δ when x ∈ V (G i ), y ∈ V (G j ), and 2 ≤ i < j ≤ k in Lemma 10, with β in place of s and (−α ; α ) ⊂ (− α 8 ; α 8 ) in place of (− α 3 ; α 3 ) as the interval of possible slopes for the edges of P x . The case in which 1 ≤ j < i ≤ k is symmetric to the previous one. Suppose now that x belongs to H and y belongs to G i , for some i ∈ {1, . . . , k}. If i = 1 and y = u 0 , then y ∈ V (H) and P xy is defined as above. Assume hence that y = u 0 . Then the path P xy consists of three sub-paths.
The first sub-path of P xy is a path P x in H from x to y 1 . Suppose first that x = u. Then let P x = τ uy1 (H). Let P x = (x = z 1 , z 2 , . . . , z s = y 1 ); we prove that d(Γ δ , z h y) > d(Γ δ , z h+1 y) holds true for any h = 1, . . . , s − 1. Consider the vertical line 1 through y 1 , oriented towards increasing y-coordinates; as argued above, the disk D ρ is to the right of 1 and y lies inside D ρ . By Properties 1 and 2 of Γ δ , the edge z h z h+1 is horizontal, with z h to the left of z h+1 . Hence, the line h orthogonal to z h z h+1 and passing through its midpoint is also vertical and has 1 to its right. It follows that y is to the right of h . Since the half-plane to the right of h represents the locus of the points of the plane that are closer to z h+1 than to z h , we have d(Γ δ , z h y) > d(Γ δ , z h+1 y). Suppose next that x = u. By Property 4 of Γ H,δ , there exists a path P x = (x = z 1 , z 2 , . . . , z s = y 1 ) in H that connects x to y 1 , that does not pass through u, and whose every edge has slope in (− α 2 ; α 2 ) in Γ H,δ . We prove that, for any h = 1, 2, . . . , s − 1, d(Γ δ , z h y) > d(Γ δ , z h+1 y); refer to Fig. 23. Since the drawing of H in Γ δ coincides with Γ H,δ , the edge z h z h+1 has slope in (− α 2 ; α 2 ) in Γ δ . Consider the line h that passes through y 1 , that is directed towards increasing y-coordinates and that is orthogonal to the line through z h and z h+1 . Denote by s h the slope of h . Then s h ∈ ( π−α 2 ; π+α 2 ). We prove that h has the disk D ρ to its right. In order to do that, consider the point p T on the half-line with slope π−α 2 starting at y 1 and such that d V (Γ δ , y 1 p T ) = ρ. Further, consider the point p B on the half-line with slope −π+α < π + s h . Now consider the line h parallel to h , passing through the midpoint of the edge z h z h+1 , and oriented towards increasing y-coordinates; h has h to its right, given that the midpoint of z h z h+1 is to the left of h in Γ δ . Thus, h has D ρ , and in particular y, to its right. Since the half-plane to the right of h represents the locus of the points of the plane that are closer to z h+1 than to z h , it follows that d(Γ δ , z h y) > d(Γ δ , z h+1 y). The second sub-path is the edge y 1 v. Since y lies in D ρ , we have that d(Γ δ , vy) ≤ ρ ≤ dy 1 v 3 . By the triangular inequality, we have that d(Γ δ , y 1 y) > d(Γ δ , y 1 v) − d(Γ δ , vy) ≥ d y1v − ρ ≥ 2dy 1 v 3 . Hence, d(Γ δ , y 1 y) > d(Γ δ , vy). The third sub-path is a path P vy that connects v to y, that belongs to k l=i G l , and that is distance-decreasing in Γ δ . This path exists, as from the case in which x and y belong to the same graph G i or from the case in which x belongs to a graph G i and y belongs to a graph G j for some 1 ≤ j < i ≤ k.
Suppose finally that x belongs to G i , for some i ∈ {1, . . . , k}, and y belongs to H. If i = 1 and x = u 0 , then x ∈ V (H) and P xy is defined as above. Assume hence that x = u 0 . We now describe the path P xy , which consists of three sub-paths.
The first sub-path of P xy is a path Q x in G i from x to u i−1 whose every edge has slope in (π − α ; π + α ) in Γ i (in Γ 1,d * if i = 1). This path exists since Γ i (Γ 1,d * if i = 1) satisfies Property 5. The second sub-path of P xy is i−1 j=1 β uj uj−1 (G j ). Since Γ j (Γ 1,d * when j = 1) satisfies Properties 1 and 2, every edge in i−1 j=1 β uj uj−1 (G j ) has slope π in Γ j (in Γ 1,d * when j = 1). Let (x = z 1 , z 2 , . . . , z s−1 , z s = y ) be the union of these two sub-paths of P xy . We prove that d(Γ δ , z h y) > d(Γ δ , z h+1 y), for any associated with a plane embedding; (b) by construction u and v are two distinct external vertices of G; (c) edge uv exists and coincides with τ uv (G), given that v immediately follows u in the clockwise order of the vertices along the outer face of G; and (d) G does not have any 2-cut, given that it is 3-connected. Thus, Theorem 7 can be applied in order to construct a planar greedy drawing of G. This concludes the proof of Theorem 1.

Conclusions
In this paper we have shown how to construct planar greedy drawings of 3-connected planar graphs. It is tempting to try to use the graph decomposition we employed in this paper for proving that 3-connected planar graphs admit convex greedy drawings. However, despite some efforts in this direction, we have not been able to modify the statement of Theorem 7 in order to guarantee the desired convexities of the angles in the drawings. Thus, proving or disproving the convex greedy embedding conjecture remains an elusive goal.