Drawing Halin-graphs with small height

In this paper, we study how to draw Halin-graphs, i.e., planar graphs that consist of a tree $T$ and a cycle among the leaves of that tree. Based on tree-drawing algorithms and the pathwidth $ pw(T) $, a well-known graph parameter, we find poly-line drawings of height at most $6pw(T)+3\in O(\log n)$. We also give an algorithm for straight-line drawings, and achieve height at most $12pw(T)+1$ for Halin-graphs, and smaller if the Halin-graph is cubic. We show that the height achieved by our algorithms is optimal in the worst case (i.e. for some Halin-graphs).


Introduction
It is well-known that every planar graph has a planar straight-line drawing in an O(n) × O(n)grid [17,24] and that an Ω(n) × Ω(n)-grid is required for some planar graphs [16] (definitions will be given in the following section).But for some subclasses of planar graphs, planar straight-line drawings of smaller area can be found.In particular, for any tree one can easily create a straight-line drawing of area O(n log n) [6]; the area can be improved to n2 O( √ log log n log log log n) [5] and O(n) if the maximum degree is O(n 1−ε ) [18].Outerplanar graphs can be drawn with area O(n 1.48 ) [7] and with area O(n log n) if the maximum degree is bounded [15] or a constant number of bends are allowed in edges [2].There are also some sub-quadratic area results for series-parallel graphs [2], though they require bends in the edges.
These existing results suggest that bounding the so-called treewidth of a graph may be helpful for obtaining better area bounds.In particular, trees have treewidth 1, and outer-planar and series-parallel graphs have treewidth 2. However, one can observe that the lower-bound graph from [16] can be modified to have treewidth 3, so we cannot hope to achieve subquadratic area for all planar graphs of constant treewidth.However, there are some subclasses of planar graphs that have treewidth 3 and a special structure that may make them amenable to be drawn with smaller area.This is the topic of the current paper.
Halin-graphs were originally introduced by Halin [20] during his study of graphs that are planar and 3-connected and minimal with this property.He showed that any such graph consists of a tree without vertices of degree 2 where a cycle has been added among the leaves of the tree.These graphs have attracted further interest in the literature, see for example [28,25,13,14,10].It is folklore that they can be recognized in linear time since they are planar graphs and have treewidth 3, but a direct and simpler approach for this was recently given by Eppstein [10].
In this paper, we study how to create planar drawings of a Halin-graph that have small area.To our knowledge, no such algorithms have been given before, and the best previous result is to apply a general-purpose planar graph drawing algorithm that achieves area O(n 2 ).In contrast to this, we exploit here that a Halin-graph consists of a tree T with a cycle C among its leaves, and give two results.The first one states that for any drawing of T , we can "fiddle in" the cycle C at a cost of increasing the height by a factor of 3.However, the resulting drawing has bends.For our second result, we take inspiration from one particular tree-drawing algorithm by Garg and Rusu [19] to create an algorithm that achieves straight-line drawings of area O(n log n).In fact, the height of our drawings, which is O(log n) in the worst case, can be bounded more tightly by O(pw(T )), where the pathwidth pw(T ) is a well-known graph-parameter.It is known that the pathwidth is a lower bound on the height of any planar graph drawing [12] and that the pathwidth of a Halin-graph is within a constant factor of the pathwidth of the tree T [13].Therefore our algorithm gives a O(1)-approximation algorithm on the height of planar straight-line drawings of Halin-graphs if we ignore small constant terms.Similarly as was done for trees by Suderman [27] and Biedl and Batzill [1], we can also argue that the constant in front of "pw(T )" cannot be improved for some Halin-graphs.Our paper is structured as follows.After reviewing the necessary background in Section 2, we briefly argue in Section 3 how to use any tree-drawing algorithm to create (poly-line) drawings of Halin-graphs of asymptotically the same height.Section 4 gives the algorithm for straight-line drawings of small height, while Section 5 defines a class of Halin-graphs that have small pathwidth, yet require a large height in any (straight-line or poly-line) planar drawing.We conclude in Section 6.

Background and notations
We assume familiarity with graphs and basic graph-theoretic terms, see for example [8].Throughout this paper, we use n for the number of vertices in a given graph G = (V, E).A tree is a connected graph without cycles.A leaf of a tree is a vertex of degree 1.A rooted tree is a tree together with one specified vertex (the root); this defines for any edge of the tree the parent-child relationship with the parent being the endpoint that is closer to the root.In a rooted tree, the term leaf is used only for those vertices that have no children, i.e., the root is not considered a leaf unless n = 1.
Fix a rooted tree T .For any vertex v ∈ T , we use T v to denote the subtree of T rooted at V , i.e., vertex v and all its descendants.We assume throughout that trees are ordered, i.e., come with a fixed cyclic order of neighbours around each vertex.In a rooted tree, this hence gives a left-to-right order of its children (starting in counter-clockwise direction after the parent).The leftmost leaf L of T is the one reached by starting at the root and repeatedly taking the leftmost child until we reach a leaf.Define the rightmost leaf R symmetrically.Note that L = R if T is a rooted path, i.e., it is a path with the root as one of its endpoints.If T consists of only one vertex (the root r), then L = r = R , but otherwise L = r = R .
Halin-graphs and skirted graphs: Let T be an (unrooted, ordered) tree without vertices of degree 2. To avoid trivialities, we assume that T has at least three leaves.Let H be the graph obtained by connecting the leaves of T in cyclical order; this is the Halin-graph formed by T (and sometimes denoted H(T )).Tree T is called the skeleton of Halin-graph H, and the edges of the cycle are called cycle-edges.See Figure 1.
Observe that any Halin-graph is planar, i.e., can be drawn without crossing in the plane.The condition 'no vertex has degree 2' is not crucial for our drawing algorithm (though it was crucial in the original study of Halin-graphs as minimal 3-connected planar graphs [20]).As in [13], we use the term extended Halin-graph for a graph H(T ) obtained by taking an arbitrary tree T and connecting its leaves in a cycle in order, while a regular Halin-graph refers to a Halin-graph as above, i.e., the skeleton has no vertices of degree 2. Our drawing algorithms will be based on rooted, rather than unrooted, trees, and therefore exploit subgraphs of Halin-graphs formed by rooted trees.Let T be an (ordered) tree that has been rooted at vertex r.Let H be the graph obtained by connecting the leaves of T in order from left to right in a path; this is the skirted graph [25] formed by T (and sometimes denoted H − (T )).Graph H − (T ) is a subgraph of H(T ); it is missing either the edge ( L , R ) or (if the root r has degree 1) the path L , r, R .
Pathwidth and rooted pathwidth: The pathwidth of a graph G is defined as follows.A path decomposition is an ordered sequence X 1 , . . ., X ξ of vertex-sets (bags) such that any vertex belongs to a non-empty subsequence of bags, and for any edge at least one bag contains both endpoints.The width of such a path decomposition is max i {|X i | − 1}, and the pathwidth pw(G) is the minimum width of a path decomposition of G.A graph consisting of a singleton vertex hence has pathwidth 0.
We will in this paper almost only be concerned with the pathwidth of trees; here an equivalent and simpler definition is known.For a path P in a tree T , let T (T, P ) denote the connected components of the graph obtained by removing the vertices of P .Suderman [27] showed that for any tree T we have where the minimum is taken over all paths P in T .Our constructions will use a rooted tree T , and therefore consider width-parameters for rooted trees that are illustrated in Figure 2. Define as in [4] the rooted pathwidth r pw(T ) as follows: where the minimum is over all rooted paths P r of T .(The recursive formula differs from the one for pathwidth only in that the path must end at the root; hence the name.)One can show that any tree T can be rooted at a leaf such that we have pw(T ) ≤ r pw(T ) ≤ 2pw(T ) + 1 [4].We call a path P r that can be used to obtain the minimum a spine.The rooted pathwidth was actually used much earlier for the classification of the order of rivers and streams [21,26] and became known as the Horton-Strahler number: where the minimum is over all children c of the root r, the maximum v is over all children of the root, and χ denotes the characteristic function.One can show [4] that the Horton-Strahler number and the rooted pathwidth are identical.We use the term spine-child for a child c where the minimum is achieved; this is the same as a child that maximizes the Horton-Strahler number among the children.(One can show that it belongs to a spine of T .) Graph drawing: A poly-line is a polygonal curve, i.e., a curve that is the union of finitely many line segments; the transition between two such segments is called a bend.A planar poly-line drawing Γ of a graph G consists of assigning a point to each vertex and an (open) poly-line to each edge such that all points and poly-lines are disjoint, and the poly-line of an edge ends at the points of the endpoints of the edge.The drawing is called y-monotone if all poly-lines of edges are y-monotone and straight-line if all poly-lines of edges are straight-line segments.
We assume throughout that identifying features (i.e., points of vertices and bends in poly-lines of edges) have integral y-coordinates.The layers of a drawing are the horizontal lines with integral y-coordinate that intersect the drawing; we usually enumerate them from top to bottom as 1, 2, . . ., h.The number h of layers is called the height of the drawing (notice that this is one unit more than the height of the minimum enclosing box).Minimizing the height of drawings is the main objective in this paper.When constructing drawings, it will sometimes be expedient to use integral x-coordinates as well; we then use the term column for a vertical line of integral x-coordinate that intersects the drawing and enumerate columns from left to right.
We usually identify the graph-theoretic object (vertex, edge) with the geometric object (point, poly-line) that corresponds to it in the drawing.All our drawings are required to be planar (i.e., without crossing edges) by definition.We often require that they are plane, i.e. reflect the given order of edges around every vertex, and (for a Halin-graph) the infinite region is adjacent to the cycle-edges.

Transforming tree drawings
In this section, we show that any order-preserving tree-drawing algorithm can be used to obtain poly-line drawings of Halin-graphs.Put differently, we can draw the skeleton-tree T , and "fiddle in" the cycle-edges.As it will turn out, we do not need to use a drawing of T ; it suffices to take a drawing of a suitably chosen subtree of T , which may make the height bound a bit smaller and (as we will see) give a tight bound.
The following defines the subtree of T that we draw; see also Figure 1.Let the inner skeleton of a Halin-graph be the tree T obtained by deleting all leaves of the skeleton.We say that T leaf-extends a tree T if T can be obtained from T by (possibly repeatedly) adding a leaf incident to a leaf of the previous tree.The leaf-reduced inner skeleton of a Halin-graph H(T ) is the smallest subgraph of the inner skeleton T that can be leaf-extended to T .We now have the following result: Theorem 1.Let H(T ) be an extended Halin-graph.If its leaf-reduced inner skeleton T has an order-preserving poly-line drawing Γ of height h, then H(T ) has a plane poly-line drawing of height 3h.
Proof. Figure 3 illustrates how to find this drawing, with the final result in Figure 1b.As a first step, insert a dummy-vertex at every bend of Γ to get a straight-line drawing Γ d of a tree T d that is tree T with some edges subdivided.Also subdivide the same edges in trees T and T (where T is the inner skeleton of H(T )) to get trees T d and T d .Next, convert Γ d into a flat visibility representation Γ vr of T d .This consists of assigning a horizontal segment s(v) to every vertex and a horizontal or vertical segment s(e) to every edge such that the segments are interior-disjoint and the segment of edge (v, w) ends at s(v) and s(w).We can always do such a conversion while giving integral y-coordinates to all segments and maintaining the same height and planar embedding [3].
We next convert visibility representation Γ vr of T d into a visibility representation Γ vr of T d .Recall that T is a leaf-extension of T , so we can obtain T d by repeatedly adding a leaf incident to a leaf p of the current tree.Since p is a leaf, there is no incident horizontal edge next to one end (left or right) of its segment s(p).We place a segment for at this end (inserting columns if needed to make space), and connect it horizontally.Repeating this gives a visibility representation Γ vr of T d .By inserting further columns, we may assume that any segment s(v) in Γ vr has at least one unit width and overhangs any incident vertical edge-segment by at least one unit.
Next, triple the grid, i.e., insert a new grid-line before and after each existing one.In consequence, we can surround the entire drawing of Γ vr with a cycle C that traces along all segments.Formally, C consists of all those points that are horizontally or vertically exactly one unit away from segments of Γ vr , and these points form a cycle since we tripled the grid.Let Γ C be the resulting drawing.Now we insert the leaves of the skeleton.Let an angle of a vertex v in T be any two consecutive edges e, e at v in T in the planar embedding.Because s(v) overhangs its incident vertical edges, cycle C has a segment s α of at least unit length for every angle α of v such that placing a leaf on s C and connecting it vertically puts edge (v, ) between e and e in the planar embedding.So for any v ∈ T and any angle α at v, insert Note that C runs within unit distance of s(e) and s(e ) at some point, and since e, e are consecutive at v, a part of C between this is within unit distance of s(v) throughout.Furthermore, since s(v) overhangs incident edge-segments, this part contains a horizontal segment s α .Insert as many leaves on s α as are required by the planar embedding of the skeleton (we can insert columns to widen s α if needed) and connect them vertically to s(v).This gives a flat orthogonal drawing Γ od : every vertex is represented by a horizontal segment, and every edge is a poly-line with only horizontal or vertical segments.Furthermore, the height is 3h and the drawing represents H(T d ) since we took care to re-insert the leaves exactly according to the planar embedding.Drawing Γ od can be converted to a poly-line drawing Γ d of H(T d ) of the same height [3].Finally by reverting dummy-vertices of T d back to bends, we obtain the desired poly-line drawing of H(T ).
Since every tree T has an order-preserving straight-line drawing Γ of height 2pw(T ) + 1 [1], we get: Corollary 1.Any extended Halin-graph H(T ) has a plane poly-line drawing of height 6pw(T ) + 3, where T is the reduced inner skeleton.
Since every tree has pathwidth at most log 3 (2n+1) [23] we can in particular draw extended Halin-graphs with height O(log n).The width can easily be seen to be O(n), so the area is O(n log n).Our construction may seem very wasteful (cycle C has many bends that could be removed with suitable post-processing stages), but as we shall see in Theorem 4, the height-bound is tight, even for some regular Halin-graphs.

Straight-line drawings
The transformation of Section 3 creates poly-line drawings, and it is not at all clear whether one could convert them into straight-line drawings without changing the height.We hence give a second, completely different algorithm that creates a straight-line plane drawing of a Halin-graph that, at the cost of doubling the height.(The width may be exponential, so this construction is of mostly theoretical interest.)Crucial for our result is that it suffices to construct poly-drawings in which all edges are drawn as y-monotone curves; by the result of Pach and Tóth [22] or Eades et al. [9] such drawings can be converted into planar drawings of the same height.
The algorithm proceeds by considering an increasingly larger subtree T of the skeleton T (rooted at an arbitrary leaf), and to draw the skirted Halin-graph H − (T ).There are three edges (called connector-edges) that connect H − (T ) with the rest of H(T ): they attach at the root and at the leftmost and rightmost leaf of T .To be able to add them later with a y-monotone curve, we restrict the locations of their endpoints.So we specify below whether the leftmost and rightmost leaf should have empty rays towards west (W) or east (E).We also restrict the root to be in the leftmost column and either as far north (N) as possible or as far south (S) as possible; sometimes either placement is acceptable and we use W to indicate this.The full set of restrictions is as as follows: Definition 1.Let T be a rooted tree with r pw(T ) ≥ 2 (and therefore L = R ).Let Γ be a plane poly-line drawing of H − (T ) in layers 1, . . ., h (enumerated top to bottom), where h ≥ 2. We call Γ an α L α r α R -drawing, for α L , α R ∈ {W, E} and α r ∈ {N, W, S}, if it satisfies the following (see also Figure 4): (d1) R is in layer 1 and L is in layer h.Root r is in the leftmost column and the only element of Γ in that column.
(d2) For X ∈ {L, R}, if α X = W, then the westward ray from X is unobstructed (i.e., intersects no other element of Γ).Otherwise (α X = E) the eastward ray from X is unobstructed.
r is in an arbitrary layer.We assumed r pw(T ) ≥ 2 in the above definition since otherwise L = R and then condition (d1) cannot be satisfied for h > 1.We hence create drawings for trees T with r pw(T ) ≥ 2 and deal with subtrees that do not satisfy this as special cases.The construction works for both regular and extended Halin-graphs, but the latter may require a bit more height.To express this succinctly, set χ ext (T ) to be 1 if T contains a degree-2 vertex that is not the root (this in particular implies that H(T ) is not regular), and χ ext (T ) = 0 otherwise.Note that χ ext (T ) ≤ χ ext (T ) for any subtree T of T .
The case r pw(T ) = 2 and some useful observations: The drawing for T if r pw(T ) = 2 is a bit special; we can save two rows (compared to drawings for higher rooted pathwidth) at the cost of no flexibility for the y-coordinate of the root.Lemma 1.Let T be a rooted ordered tree with r pw(T ) = 2. Then for any α Proof.See Figure 5a for the following construction.Fix a spine P that goes from root to a leaf, and place P on one layer, with the root leftmost.
Any T ∈ T (T, P ) has rooted pathwidth 1 since P is a spine.If χ ext (T ) = 0, then T has no vertices of degree 2, so it is a single leaf.Place it in the layer above or below P depending on whether T is right or left of the spine P .The cycle-edges can now be completed along these layers.If χ ext (T ) = 1, then initially contract all vertices of degree 2 and draw the tree as above.Then insert extra layers before/after the spine-layer and place degree-2 vertices (or a bend, if there are none) within those layers.
So we have constructed a WWW-drawing of height 3 + 2χ ext (T ).Any of the other drawing-types is constructed by "turning rays" around.We describe this in a more general lemma below since it will be useful for later cases as well.
Proof.Leaf R is in the topmost layer, so its incident edges are routed y-monotonically and leave horizontally or downward from R .To achieve α R = E, add a new layer above Γ, move R into it, and extend its incident edges via a bend near where R used to be.See also Figure 5b.This gives a y-monotone drawing where the bottom layer is unchanged (in particular, it still contains L with its unobstructed ray).The root is no longer be in N-position if it was before, but this is not a problem since we only promised an α L Wα R -drawing.Similarly one achieves α L = E by adding a layer below Γ and moving L into it.
The following will be useful when merging a drawing of a subtree T that uses fewer layers than permitted (e.g. because χ ext (T ) = 0 while χ ext (T ) = 1): We can "pad" such a drawing by inserting empty layers suitably, even while maintaining the drawing type.
Claim 2. Assume that H − (T ) has a y-monotone α L α r α R -drawing of height h ≥ 3. Then for any h > h it also has a y-monotone α L α r α R -drawing of height h .
Proof.First insert bends whenever an edge crosses a layer without a bend; now all edgesegments are horizontal or connect adjacent layers.If α r = N then r is in layer 2. Insert h − h horizontal grid-lines between layer 2 and layer 3, and add bends to any edge that crosses the inserted lines.So edge-segments again are horizontal or connect adjacent grid-lines, which means that we can change the y-coordinates of grid-lines to be integers (i.e., stretch the drawing between layers 2 and 3) without affecting planarity or y-monotonicity.This gives the desired α L Nα R -drawing since r remains in layer 2, and no changes were made within the top or bottom layer.The construction is symmetric (inserting layers between h − 2 and h − 1) for α r = S, and either construction can be used for α r = W.
The induction hypothesis: We create drawings for arbitrarily large rooted pathwidth by induction; the following states the induction hypothesis.(It differs from Lemma 1 in that we sometimes permit α r = N or α r = S while Lemma 1 only holds for α r = W.) Lemma 2. Let T be a rooted ordered tree with r pw(T ) ≥ 3, and let α L , α r , α R be any of the combinations WWE, EWW, EWE, WNW and WSW.Then H − (T ) has a plane y-monotone α L α r α R -drawing of height 6r pw(T Before proving this lemma, we briefly argue why it suffices.Specifically, the height is 6r pw(T ) − 9 + 2χ ext (T ) for a suitable choice of root for T .
Proof.Root the skeleton T at a leaf r such that r pw(T ) ≤ 2pw(T ) + 1 [4].Apply Lemma 1 or 2 to this rooted version of T to obtain a y-monotone WWW-drawing of H − (T ) of height 6r pw(T ) − 9 + 2χ ext (T ) ≤ 12pw(T ) − 3 + 2χ ext (T ).The westward ray from L is unobstructed; we can draw ( L , r) along this ray until the leftmost column and then go up to r. Likewise we can draw ( R , r) to obtain a y-monotone drawing of H(T ).This can be transformed into a straight-line drawing of the same height [22,9].
The height of Theorem 2 is (roughly) a factor 2 worse than the height in Corollary 1.However, in terms of rooted pathwidth, Theorem 2 is tight, see Theorem 5 and 6.
The rest of this section is dedicated to the proof of Lemma 2. It suffices to show how to construct a WNW-drawing of height h := 6r pw(T ) − 9 + 2χ ext (T ) ≥ 9; the construction of a WSW-drawing is symmetric and all other cases are covered by Claim 1.
We use the following notations throughout.Let r be the root of T , let d be its degree, and let c 1 , . . ., c d be the children of the root, in order.We use the notation i L and i R (for i = 1, . . ., d) for the leftmost and rightmost leaf of T c i .Recall that HS(T ) = r pw(T ) ≥ 3. Let c s be the spine-child of the root; by definition of Horton-Strahler number this is the only child whose subtree could have the same rooted pathwidth as T .If r pw(T cs ) < r pw(T cs ) then (to avoid some cases) we re-assign s = d.Whether or not we reassigned, we hence have r pw(T c i ) < r pw(T ) for all i = s.
We prove the lemma by induction on r pw(T ), with the base case at r pw(T ) = 3.We do an inner induction on the size of the tree, and use as base case the case r pw(T cs ) < 3 (this must occur since at the leaf of the spine the rooted pathwidth is 1).Much of the construction will be the same for base case and induction step, and we therefore prove them together.

Drawing subtrees up to T cs :
The following algorithm (illustrated in Figure 6) states which drawing to use for each subtree and how to combine them.We build the drawing left-to-right, beginning with the root and then adding the subtrees at the children.
Figure 6: The constructions if c s is the rightmost child.
1. Place the root r in the leftmost column of layer 2. We reserve the eastward ray from r for edge (r, c d ), i.e., we will make sure that nothing is added to intersect it until the edge-segment is completed.
2. For i = 1, . . ., d−1, if r pw(T c i ) ≥ 2, then recursively (or via Lemma 1) obtain an WWEdrawing Γ i .This has height at most 6r pw( If needed we can use Claim 2 to make Γ i have height exactly h − 5. Place Γ i in layers 6, . . ., h, to the right of everything drawn thus far. If r pw(T c i ) = 1, then T i is a rooted path.Place its leaf on layer h and its degree-2 vertices (if any) on layer h − 1, with the root leftmost.
We place (parts of) the connector-edges of T c i as follows: • Connect c i to r by going upward to layer 3 and then (via a bend) to layer 2.
• We draw part of the connector-edge ( i R , i+1 L ) by going eastward from i R (in its layer) beyond Γ i , and adding (if needed) a bend to go downward to layer h.The eastward ray in layer h from here is reserved for edge ( i R , i+1 L ).
L is the leftmost leaf of T ; its westward ray is unobstructed as required.For i > 1, leaf i L was placed on the ray reserved for edge ( i−1 R , i L ), which is hence completed.Since this edge receives no further bend at i L , and was drawn y-monotonically extending from i−1 R , it is drawn y-monotone.

3.
To handle the spine-child c s we have three cases.
Assume first that s = d and r pw(T cs ) ≥ 2. Recursively (or via Lemma 1) obtain a WWW-drawing Γ s of H − (T cs ) and increase its height (if needed) to be h.Place Γ s in layers 1, . . ., h, to the right of everything drawn thus far.Connect c s to r by going upward to layer 2 and then horizontally to r. Edge ( s−1 R , s L ) is completed automatically, and s R is the rightmost leaf and its eastward ray is unobstructed.Assume next that s = d and r pw(T cs ) = 1.(This can happen if we re-assigned s.) Place the leaf of T cs on layer 1 and all other vertices on layer 2, with the root leftmost.(If |T cs | = 1 then place a bend in row 2.) Edge (r, c s ) is completed automatically, and s R has an eastward unobstructed ray.To route connector-edge ( s−1 R , s L ), we undo the partial routing that we did earlier; instead we go eastward from d−1 R and then upward to d L = d R in row 1. Assume finally that s < d, i.e., c s is not the rightmost child.The drawing here is much more complicated and will be explained below.
Drawing the remaining subtrees if s < d − 2: Our construction is done if s = d, so assume not.By the re-assignment, this implies r pw(T cs ) = r pw(T ).In particular, we are not in the base case of the inner induction, and we know that r pw(T cs ) ≥ 3.This allows us (crucially) to choose an WSW-drawing for H − (T cs ), which in turn permits us to route (r, c s ) y-monotonically while leaving sufficiently much space for T c s+1 , . . ., T c d .
We assume for now that s ≤ d − 2; the case s = d has been dealt with above and the case s = d − 1 is not difficult but requires a variation that will be explained below.
reserved for (r, cs) 4. Draw parts of the edge (r, c s ), by going from r to a bend in layer 3 (to the right of everything drawn thus far), then down to another bend in layer h − 1.We reserve the eastward ray in layer h − 1 from this bend for edge (r, c s ).

5
. By s ≤ d − 2 child c s+1 exists and is not c d .
If r pw(T c s+1 ) ≥ 2, then let Γ s+1 be a recursively obtained EWE-drawing of T c s+1 ; since r pw(T c s+1 ) < r pw(T ) this has height at most h − 4. Increase its height (if needed) so that it has height exactly h − 4, and place Γ s+1 in layers 3, . . ., h − 2 to the right of everything drawn thus far.If T c s+1 is a rooted path, then place all its vertices in layer h − 2. Draw the connector-edges as follows: • Connect c s+1 to r by going upward to layer 3 and then (via a bend) to layer 2.
• Leaf s+1 L is placed in layer h − 2; we reserve its eastward ray for edge ( s+1 L , s R ). • If T c s+1 is not a rooted path, then leaf s+1 R has an unobstructed eastward ray; we begin drawing edge ( s+1 R , s+2 L ) by going eastward from s+1 R , then vertically to layer h − 3 and reserving the eastward ray in layer h − 3 for ( s+1 R , s+2 L ).If T c s+1 is a rooted path, then s+1 R is in layer h − 2. We go up one unit to layer h − 3 and reserve the eastward ray for ( s+1 L , s R ).
6.For i = s + 2, s + 3, . . ., d − 1, we process T c i and its connector-edges as we did in Step 2, only we put the drawing three levels higher.
was placed in layer h − 2, and the edge was routed by going upward to layer h − 3. We now go eastward from there and then upward to layer 1.This is the only situation where a connector-edge receives bends when placing both endpoints, but one verifies that this route is y-monotone.

Recall from
Step 5 that the eastward ray in layer h − 2 was reserved for connector-edge ( s+1 L , s R ).We now add a bend in it to the right of everything drawn thus far, then go vertically to layer 1 and reserve the eastward ray.9. Finally, recursively obtain an WSW-drawing Γ s of T cs of height h.(This exists since c s is not the rightmost child, hence r pw(T cs ) = r pw(T ) ≥ 3 and induction can be applied.)Place Γ s to the right of everything drawn thus far.Since s R , c s , s L are in layers 1, h − 1 and h, respectively, this completes the connector-edges of T cs .
The case s = d − 1: Previously, we used an EWE-drawing for c s+1 and an WNW-drawing for c d in Steps 5 and 7.If s = d − 1, then c s+1 = c d takes on the roles of both of these drawings.The following step (see Figure 7) replaces step 5 and 6 if s = d − 1.We have constructed a WNW-drawing in all cases, and one easily verifies that all edges are drawn y-monotonically, hence Lemma 2 and with it Theorem 2 holds.
It is worth mentioning that this poly-line drawing can easily be found in linear time, as long as coordinates of vertices are expressed initially with via offsets to their parents, and evaluated to their final value only after finishing the construction of the entire tree.

Halin-graphs with maximum degree 3
Observe that in Figures 6 and 8 (where s ∈ {d − 1, d}) we are "wasting" layers; the same construction could have been done with three fewer layers.This leads to the following.Lemma 3. Let T be a rooted binary tree with r pw(T ) ≥ 2, and let α L , α r , α R be any of the combinations WWE, EWW, EWE, WNW and WSW.Then H − (T ) has a plane y-monotone α L α r α R -drawing of height 3r pw(T Proof.We again proceed by induction and show that there exists an WNW-drawing of height 3r pw(T ) − 2 (all other drawing-types are symmetric or obtained with Claim 1).We only sketch the necessary changes to the previous algorithm here; the reader should be able to fill in the details using Figure 9.The previous base case construction gives 3 + 2χ ext (T ) layers.We can also achieve at most 4 layers, by placing a spine-vertex on layer 2 if the spine-child is the right child and on layer 3 otherwise.Using the better of the two (depending on χ ext (T )) we hence have 3 + χ ext (T ) = 3r pw(T ) − 3 + χ ext (T ) layers.In the induction step, we have d ≤ 2 children and hence always either s = d or s = d − 1.So construct a drawing of H − (T ) as in Figure 6 or Figure 8, except use h = 3r pw(T ) − 3 + χ ext (T ) and place drawing Γ i for i < s in layers 3, . . ., h.
If an extended Halin-graph has maximum degree 3, then its skeleton T is binary when rooting it at a leaf.Since we can do so and achieve r pw(T ) ≤ 2pw(T ) + 1, this implies as in the proof of Theorem 2: Theorem 3. Every extended Halin-graph with maximum degree 3 and skeleton T has a straight-line drawing of height at most 6pw(T ) + χ ext (T ).

Lower bounds on the height
Both papers that gave approximation algorithms for the height on tree drawings [27,1] also constructed trees where this bound is tight.In particular, Batzill and Biedl showed that there exists an ordered tree that requires height 2pw(T ) + 1 in any ordered drawing [1].In the same spirit, we now construct Halin-graphs that need as much height as we achieve with our algorithms. 1  Definition 2. For w ≥ 1, define C w and F w as follows: • C 1 consists of a path r, c (where r is the root) with a leaf attached at each of them on each side of the path.See Figure 10a.
• F w is obtained from C w as follows.Let r be the root of C w .Add a parent p and a grand-parent g to r and make g the root.Attach a leaf on each side of path p, g, r at each of p, g.See Figure 10b. 1 The graphs were chosen as to keep the argument as simple as possible; like much smaller trees would do.
• C w+1 is obtained as follows.Start with a spine consisting of vertices s 1 , . . ., s S for some sufficiently large constant S that we will specify later, and make s 1 the root.At each spine-vertex except s S , attach on each side of the spine L copies of F w via its root, for some sufficiently large constant L that we will specify later.See Figure 10c.We prove Lemma 4 by induction on w.In the base case (w = 1) vertex c in C 1 is surrounded by a 5-cycle in H − (C 1 ).Since we need one layer for c, and two more layers to surround it, any plane drawing of H − (C 1 ) requires three layers as desired.The induction step will proved over the next four subsections, but we sketch here the main idea.Fix an arbitrary plane poly-line drawing Γ of H − (C w+1 ) for some w ≥ 1. Tree C w+1 contains lots of copies of F w , hence of C w .Therefore, Γ contains lots of copies of H − (C w ); each of them uses at least h(w) layers by induction.We can argue that some copy of H − (F w ) inside Γ actually requires h(w) + 1 layers; this is the most difficult part that we defer to last.Furthermore, there are 5 polylines inside Γ that are disjoint from this copy of H − (F w ) and that "bypass" it (defined below).It is known that 5 bypassing polylines need 5 additional layers.Therefore the height is at least h(w) + 1 + 5 ≥ h(w + 1).

Preliminaries and preprocessing
We first introduce some terms concerning the abstract tree C w+1 .Recall that C w+1 is rooted and has a total order among the children of every vertex.We therefore have a total order among the leaves, starting at the leftmost leaf and ending with the rightmost one.However, we will use "left" to refer to the order of vertices within one level of the drawing, which may or may not reflect the order in the tree.To avoid confusion, we will therefore treat the order of chidren/leaves as if it were time, and so speak of the "first"/"last" leaf and that a leaf comes earlier than another.
We distinguish leaves of C w+1 (other than s S ) by whether they are on the before-spine or after-spine, i.e., before or after s S in the enumeration of leaves.Likewise for a spine-vertex s i = s S we distinguish the non-spine children by whether they are before or after the spine.Any such non-spine child g is the root of a copy of F w which we denote by F (g).For any two leaves , of C w+1 , the cycle-path from to consists of the subpath of the cycle-edges between and .Now we introduce some terms concerning drawing Γ. Enumerate the layers of Γ, from top to bottom, as 1, 2, . . ., h.We are done if h ≥ h(w + 1), so assume for contradiction that h < h(w + 1) = h(w) + 6.In fact we may assume h = h(w) + 5 because we can add empty layers.For two points p, q, we write p ≺ q (or "p is left of q") if p and q are on the same layer and p has smaller x-coordinate.
A few minor modifications to drawing Γ will make later arguments easier and do not affect the height.First, insert a bend into any edge-segment that crosses a layer without having a bend there.(These new bends may not have integral x-coordinates, but integrality of x-coordinate is never used in the lower-bound proof.)Second, do the following for any spine-vertex s i (with i < S) of C w+1 , and any non-spine child g of s i .Recall that g had three children; one is vertex p while two are leaves.Delete the two edges to these leaves; their sole purpose was to ensure that the Halin-graph is regular and they will not be used in the proof.With this, g now has degree 2. For the third modification, if (s i , g) is not drawn as a straight-line, then move g to the bend on (s i , g) nearest to s i .This makes (s i , g) a straight-line and (by the first preprocessing step) puts g either on the same level as s i or one level above or below; this will be frequently used below.
Recall that F (g) denotes the copy of F w attached at g.We use Γ(g) for the drawing of H − (F (g)) as it appears after these modifications.Since Γ(g) contains a drawing of H − (C w ) within, it must use at least h(w) layers.
Finally we briefly review the concept of bypassing (see also Figure 11a); we use a version here that is 90 • rotated from the one in [4].Recall that bends of a polyline (like all bends and vertices of Γ) are required to have integral y-coordinates.Definition 3. Consider a set of poly-lines π 1 , . . ., π k that are disjoint except (perhaps) at their endpoints.Let π be a poly-line that is disjoint from π 1 , . . ., π k .We say that π 1 , . . ., π k bypass π if there exists a layer that intersects π, and for i = 1, . . ., k poly-line π i begins and ends in layer and all points in π ∩ are between the two ends of π i .Lemma 5. [4] If a planar poly-line drawing Γ contains k poly-lines that bypass a poly-line π, and if π intersects h layers, then Γ uses at least h + k layers.

The ideal case
We first argue that the height-bound holds in one special case; we will show later that this situation must occur somewhere in C w+1 (up to symmetry), as long as S and L are big enough.We assume that the following holds (see also Figure 11b): π F (C1) There are three spine-vertices s j 1 , s j 2 , s j 3 that are all located in one layer ≤ 5. Furthermore, 1 ≤ j 1 < j 2 < j 3 < S and s j 1 ≺ s j 2 ≺ s j 3 .
(C2) For k = 1, 2, 3, vertex s j k has an after-spine child g j k and a before-spine child g j k on layer +1.In fact, s j 2 has five after-spine children on layer +1.
Furthermore, one of the spine-edges incident to s j 2 has a bend or endpoint b on layer + 1.If b is on edge (s j 2 , s j 2 −1 ) then g (3) ≺ b, otherwise b ≺ g (1) .
We will later argue that the following property holds automatically, given (C1-C4).
(C5) There exists a path π within Γ(g (2) ) that connects g (2) (which is on layer + 1) to layer + h(w) + 1, and all points in π ∩ ( + 1) lie strictly between g (1) and g (3) .Now we define five interior-disjoint paths in C w+1 as follows: (see also Figure 11c): • π 1 : This path begins at g j 1 , continues within F (g j 1 ) to the last leaf, and from there along the cycle-path to the first leaf of F (g j 3 ).From there it goes upwards in the tree to g j 3 .This path uses only F (g j 1 ) and F (g j 3 ) and cycle-edges among leaves that are before the spine.
• π 2 : This path begins at g j 3 , continues within F (g j 3 ) to the last leaf, and from there along the cycle-path to the first leaf of F (g (1) ).From there it goes upwards in the tree to g (1) .This path uses only F (g j 3 ) and F (g (1) ) and cycle-edges among leaves that are between s S and the first leaf of F (g (1) ) in the total order of leaves.
• π 4 : This path is built symmetrically to π 2 : begin at g j 1 , go to the first leaf of F (g j 1 ), from there along the cycle-path (in reverse) to the last leaf of F (g (3) ), and from there to g (3) .This path uses only F (g j 1 ) and F (g (3) ) and cycle-edges among leaves that the last leaf of F (g (3) ) or later.
• π 5 : Recall that one bend b of a spine-edge incident to s j 2 lies on layer + 1. Path π 5 begins at b, and goes along spine-edges, away from s j 2 , until it reaches either s j 1 or s j 3 .
From there it goes to the after-spine child on layer + 1, i.e., either g j 1 or g j 3 .Except for this last edge, π 5 uses only spine-edges.
Proof.Directly from the edges that they use, one observes that the five paths are disjoint from π, and from each other except that they may have endpoints in common.(We use here that g (2) lies between g (1) and g (3) in the order of children at s j 2 by (C3).)Assume that b is right of g (3) , the other case is symmetric.Then all five paths begin at a point in {g j 1 , g j 1 , g (1) } and end at a point in {g (3) , b, g j 3 , g j 3 }.Observe that g j 1 is necessarily left of g (1) , otherwise the straight-line segments (s j 1 , g j 1 ) and (s j 2 , g (1) ) would intersect.Likewise g j 1 ≺ g (1) and g (3) ≺ g j 3 , g j 3 .So all five paths connect a point on layer + 1 that is at or to the left of g (1) with a point on layer + 1 that is at or to the right of g (3) .Since π uses only points on ( + 1) that are strictly between g (1) and g (3) by (C5), the claim holds.

Guaranteeing conditions (C1-C4)
Now we argue that conditions (C1-C4) are satisfied at some subtrees if S and L are big enough.Recall that we assumed (for contradiction) that h = h(w) + 5. Since each copy of H − (F w ) uses at least h(w) layers, we therefore have only 5 layers for bypassing any copy of H − (F w ).Roughly speaking, this forces spine-vertices to be in the top 5 or the bottom 5 layers.Therefore (C1) holds if S is big enough.Next we argue that of the L attached copies of F w at a spine-vertex s, only L − 72 can share a layer with s.This, plus the preprocessing, forces (C2) if L ≥ 81.It also implies that many non-spine children satisfy (C4), and an appropriate choice among them ensures (C3).
To give the details, we first study various properties of non-spine children of one fixed spine-vertex s i with i < S.
Proof.There are h = h(w) + 5 layers in total, and by induction Γ(g) intersects at least h(w) layers.It therefore can avoid only the top 5 and the bottom 5 layers.
We say that g is bad if the layer of s i intersects Γ(g), otherwise g is good.
Claim 4. At most 72 non-spine children of s i are bad. 2roof.We say that a non-spine child g has type (t, b) if the topmost and bottommost layer used by Γ(g) are t and b.By Observation 1 we have 1 ≤ t ≤ 6 and h(w) ≤ b ≤ h(w) + 5, so there are at most 36 types.Assume for contradiction that there are 73 = 2 • 36 + 1 bad non-spine children of s i , hence three of them (say g 1 , g 2 , g 3 ) have the same type (t, b).
For k = 1, 2, 3, let B k be a poly-line within Γ(g k ) that begins in layer t and ends in layer b.Let Q k be a poly-line that starts at s i (which is within layers {t, . . ., b} since g k is bad), goes along the straight-line edge to g k (also within {t, . . ., b}) and continues within Γ(g k ) until it reaches B k .Note that B 1 ∪ Q 1 and B 2 ∪ Q 2 and B 3 ∪ Q 3 are disjoint except at s i , and reside entirely within layers {t, . . ., b}.See also Figure 12a.
Exactly as in the proof of Lemma 5 in [1], one argues that this is impossible.Consider the drawing induced by k (B k ∪ Q k ).Add a vertex v in layer t − 1 and connect it to the top ends of B 1 , B 2 , B 3 (they are in layer t).Likewise add a vertex v in layer b + 1 and connect it to the bottom ends of B 1 , B 2 , B 3 (they are in layer b).This gives a planar drawing of K 3,3 , with {s i , v , v } as one side and the points B k ∩ Q k for k = 1, 2, 3 as the other side.Contradiction.Corollary 2. If L ≥ 37 then the layer of s i is in {1, . . ., 5} ∪ {h(w)+1, . . ., h(w)+5}. 3roof.If s i were in any layer in {6, . . ., h(w)}, then by Observation 1 all 2L ≥ 74 non-spine children of s i would be bad.
Claim 5.If s i is on layer where ≤ 5 and ≤ h/2, and if L ≥ 81, then s i has at least 5 good after-spine children on layer + 1.
Proof.There are L after-spine children, hence at least L − 72 ≥ 9 that are good.Any such good child g cannot be on layer by definition of good, and it is at most one layer away by the preprocessing.So g is on layer − 1 or + 1. Assume for contradiction that there at most 4 good after-spine children on layer + 1.So at least 5 good after-spine children are on layer − 1, call them g 1 , . . ., g 5 , enumerated in left-to-right order along the layer.We now have two cases.In the first case, ≤ h(w) (which is always true for w ≥ 2 since then h(w) ≥ 9 while ≤ 5).Since g 1 is good, drawing Γ(g 1 ) cannot use layer , so it is contained within layer 1, . . ., − 1.So it uses at most h(w) − 1 layers, which is impossible.
Since edge (s i , g k ) (for k = 1, . . ., 5) is drawn straight-line by the pre-processing, and Γ respects the planar embedding, the cyclic order of neighbours of s i must contain g 1 , . . ., g 5 in this order.The spine-edges and before-spine children at s i may appear somewhere between g 1 and g 5 in the cyclic order, but regardless of where they are, either g 1 , g 2 , g 3 or g 3 , g 4 , g 5 are a subsequence of the linear order of children of s i .By Claim 6 (proved below, but there is no circularity) drawing Γ(g 2 ) or Γ(g 4 ) hence uses a point on layer + h(w) + 1 = 5 + 3 + 1 = 9.This gives the required contradiction of our assumption.Now we explain how to satisfy (C1)-(C4).Assuming S ≥ 42, we have 41 spine-vertices s i with i < S. Assuming L ≥ 37, each of them is on one of 10 possible layers by Corollary 2. By the pigeon-hole principle, therefore, at least 5 of these spine-vertices are on one layer .After a possible vertical flip of Γ, we may assume ≤ h/2, therefore ≤ 5 by Corollary 2. 4 Among the 5 spine-vertices on , we can (by the Erdős-Szekeres theorem [11]) find a subsequence of √ 5 = 3 spine-vertices s j 1 , s j 2 , s j 1 such that j 1 < j 2 < j 3 and either s j 1 ≺ s j 2 ≺ s j 3 or s j 3 ≺ s j 2 ≺ s j 1 .After a possible horizontal flip of Γ we have s j 1 ≺ s j 2 ≺ s j 3 and therefore (C1) holds.
(C2) holds (assuming L ≥ 81) due to Claim 5 and a symmetric lemma, proved exactly the same way, for before-spine children.
To argue (C3), let g 1 , . . ., g 5 be the 5 after-spine children of s j 2 that are good and on layer + 1, enumerated in left-to-right order along the layer.Let g be a before-spine child of s j 2 that is on layer + 1, and notice that the cyclic order of neighbours of s j 2 contains g , s j 2 +1 , g 1 , . . ., g 5 , s j 2 −1 =: ρ as subsequence.Since the edges from s j 2 to g , g 1 , . . ., g 5 are straight-line by the pre-processing, the x-coordinate order of g , g 1 , . . ., g 5 along layer + 1 must fit the (cyclic) order ρ.Depending on whether g 3 is right or left of g , therefore either g ≺ g 1 ≺ g 2 ≺ g 3 or g 3 ≺ g 4 ≺ g 5 ≺ g , See Figure 12b.

Arguing (C5)
So we have now found subtrees such that (C1-C4) hold.This always implies (C5), but the argument for this is lengthy.We also need to prove the missing piece for Claim 5.Both will be done with the same argument as follows.Claim 6.Let s i (for i < S) be a spine-vertex on layer that has three good after-spine children g (1) , g (2) , g (3) on layer + 1 and the order of children at s i contains g (1) , g (2) , g (3) as subsequence.Then there exists a path π within Γ(g (2) ) that connects g (2) to layer + h(w) + 1, and all points in π ∩ ( + 1) lie between g (1) and g (3) .Proof.Recall that tree F w is built by extending tree C w ; let C be the copy of C w that is inside F (g (2) ).Also let I be the open interval of points on layer + 1 between g (1) and g (3) , so path π should intersects layer + 1 only in I.We need an observation.
Observation 2. H − (C) uses no points in I.
Proof.Define a cycle Q in H − (C w+1 ) as follows.Start at the unique child p of g (2) , go to its last child R (which is a leaf) and from there along the cycle-path to the first leaf of F (g (3) ).Go upwards in tree F (g (3) ) to g (3) and from there to s i .Continue symmetrically through F (1) , i.e., go from s i to g (1) to the last leaf of F (g (1) ), then along the cycle-path to the first child L of p and then to p. See Figure 13a.This cycle separates g (2) from H − (C) in the planar embedding since g (2) is between g (1) and g (3) in the order of children of s i .Now study the corresponding poly-line Q in Γ.Since g (1) , s i , g (3) is drawn with straightline segments between layers + 1 and , and since g (2) ∈ I and Γ is plane, all of I is on or inside Q.On the other hand H − (C) is strictly outside Q and the claim holds.
Let the pocket P be defined as follows, see also Figure 13b.For k = 1, 3, let B k be a poly-line within Γ(g (k) ) that connects g (k) to a point b k on layer + h(w); this exists since Γ(g (k) ) spans at least h(w) layers and contains no point in layer .We choose b k such that B k is minimal, i.e., contains no other point on layer + h(w); in particular all its points are hence in layers + 1, . . ., + h(w).Let the lid σ be the line-segment b 1 b 3 ; note g (1)   s (a) (b) that σ is not necessarily a segment of Γ.Now define pocket P to be the set bounded by B 1 ∪ g (1) , s i , g (3) ∪ B 3 ∪ σ, where the lid σ is included in P while all other points on the boundary are excluded.Note that any point in ( + 1) ∩ P is in I, because B 1 and B 3 contain no points on layer or above by (C4).Assume for contradiction that all of Γ(g (2) ) (and in particular therefore H − (C)) resides within pocket P .Then H − (C) uses no points on layer + 1, because it does not use points in I. Therefore H − (C) fits within h(w) − 1 layers, a contradiction.So Γ(g (2) ) must use points outside the pocket.These cannot be on B 1 ∪ B 3 or g (1) , s i , g (3) since these paths do not belong to F (g (2) ).So to get to a point outside P , some polyline of Γ(g (2) ) must contain a point q on σ ⊂ P from which it goes downward.Let q be the next bend of this polyline, which is on layer + h(w) + 1 by the preprocessing.Let π be the poly-line from g (2) (on layer + 1) to point q (on layer + h(w) + 1) that is within Γ(g (2) ).With the exception of the segment from q to q, poly-line π was inside pocket P ; in particular it can use no points on layer + 1 except the ones that are on I.This proves the claim.So we have proved Claim 6, which finishes the proof of Claim 5. Hence (C2) holds.From this we derived (C3) and (C4), hence the precondition for Claim 6 holds for the three children g (1) , g (2) , g (3) of s j 2 that we chose.Claim 6 hence implies (C5) and the proof of Lemma 4 is complete.

Proving the lower bounds
We now finally prove the lower bounds.To do so, we first bound the (rooted) pathwidth of F w and trees derived from it.Observation 3. We have r pw(F w ) ≤ w+1 and pw(F w ) ≤ w−1, where F w is the leaf-reduced inner skeleton of H(F w ).
Proof.We proceed by induction on w.Tree F 1 consists of a path g, p, r, c with leaves attached; this has rooted pathwidth 2. Also F 1 consists only of g, since it is obtained from F 1 by first deleting all leaves (this gives a path), and then repeatedly doing leaf-reductions (this removes all but g).So pw(F 1 ) = 0. Now consider F w+1 for w ≥ 1.This consists of a path g, p, s 1 , . . ., s S with copies of F w attached.Using this path as spine, we immediately get r pw(F w+1 ) ≤ r pw(F w ) + 1 ≤ w + 2. Also, F w+1 consists of the same path with copies of F w attached; therefore pw(F w+1 ) ≤ pw(F w ) + 1 ≤ w.
Thus far all constructions and lower bounds have been for plane drawings (respecting the embedding and have the cycle-edges at the infinite region).But we can easily prove lower bounds even for planar drawings which have no requirement except to be crossing-free.Theorem 4.There exists a regular Halin-graph H(T ) such that any planar poly-line drawing of H(T ) requires at least 6pw(T ) + 3 layers, where T is the reduced tree of the inner skeleton of H(T ).
Proof.For any w ≥ 2, consider the tree T obtained by taking two copies of F w and combining them by adding an edge between the two copies of the root g.Fix an arbitrary planar poly-line drawing Γ of H(T ).Since H(T ) is 3-connected [20] the clockwise order of edges must be the same in H(T ) and in Γ.But the infinite region of Γ could be incident to some face different from the one bounded by the cycle-edges.Tree T contains two copies of F w , and the infinite region of Γ can be a face of H − (F w ) for at most one of them.Therefore Γ contains a plane drawing of H − (F w ), hence also one of H − (C w ).By Lemma 4 this requires at least h(w) = 6w − 3 layers.The reduced inner skeleton of H(T ) consists of two copies of F w , each of which had pathwidth at most w − 1, and this bound is obtained with a main path that ends at g. Therefore we can use the two combined paths as main path for T and so have pw(T ) ≤ w − 1 and the bound holds.
We note that this lower bound implies a lower bound of Ω(log n) on the height, since C w contains c w vertices for some (rather large) constant c.However, this bound is not new since already using the Halin-graph of a complete ternary tree could give a lower bound of Ω(log n) on the height.The main contribution of our lower bound is that it matches the upper bound relative to "pw(T )" in Theorem 1. (This was also the reason why we used the leaf-reduced inner skeleton, rather than the skeleton, in Theorem 1.) We also promised a lower bound in terms of the rooted pathwidth.Note that the skeleton of a Halin-graph is an unrooted tree T ; to be able to talk about r pw(T ) we define this to be the minimum over all choices of the root.Theorem 5.There exists a regular Halin-graph H(T ) such that any planar poly-line drawing of H(T ) requires at least 6r pw(T ) − 9 layers.Proof.For any w ≥ 2, again let T be two copies of F w , combined by adding an edge between the two roots.We know r pw(F w ) ≤ w + 1, and the same holds for T if we root it suitably.Namely, the spine of F w is g-p-s 1 -. . .-s S ; if we root T at one copy of s S then we can use as its spine the two combined spines of the two copies of F w and have the same rooted pathwidth.H(T ) is a regular Halin-graph and since (as above) any planar drawing of it includes a plane drawing of H − (C w ), by Lemma 4 it requires at least h(w) = 6w − 3 ≥ 6r pw(T ) − 9 layers.

Figure 1 :
Figure 1: (a) A regular Halin-graph.Cycle-edges are blue dashed/dotted, skeleton T is black/gray, and the skirted graph H − (T ) would omit the dotted edge if T were rooted at r.The inner skeleton is gray, the leaf-reduced inner skeleton is light gray.(b) A poly-line drawing obtained with the transformation in Section 3.

Figure 2 :
Figure 2: The skeleton-tree T of Figure 1 has HS(T ) = r pw(T ) = 3. Numbers indicate the Horton-Strahler number; thick paths (solid red for the whole tree, dashed blue for the subtrees) are possible spines.

Figure 3 :
Figure 3: Transform (a) a poly-line drawing of the leaf-reduced inner skeleton (with a white dummy-vertex inserted at a bend) into (b) a visibility represention.(c) Expand leaves and widen vertex-segments to overhang (x-coordinates are not to scale).Then (d) triple the grid and insert cycle C and the leaves to get a flat orthogonal drawing (inserted columns are not shown).

Theorem 2 .
Every regular Halin-graph H(T ) has a straight-line drawing of height at most 12pw(T ) − 3, and every extended Halin-graph H(T ) has a straight-line drawing of height at most 12pw(T ) − 1.

Figure 7 :
Figure 7: (Top) The construction if s ≤ d − 2. (Bottom) The construction again, with other subtrees as rooted paths.

7 . 1 R
We process T c d very similarly to Step 3.So assume first that r pw(T c d ) ≥ 2. Recursively (or via Lemma 1) obtain an WWWdrawing Γ d of H − (T c d ) of height at most h − 6. Increase its height to be h − 3. Place Γ d in layers 1, . . ., h − 3, to the right of everything drawn thus far.Connect c d to r by going upward to layer 2 and then horizontally to r. Edge ( d−1 R , d L ) is completed automatically, and d R is the rightmost leaf and its eastward ray is unobstructed.Now assume that T c d is a rooted path.Place the leaf of T c d on layer 1 and all other vertices on layer 2, with the root leftmost (if |T cs | = 1, then place a bend in row 2).Edge (r, c s ) is completed automatically, and d R (which is the rightmost leaf) has an eastward unobstructed ray.To route connector-edge ( d−1 R , d L ), we have two cases.If d > s + 2 and/or r pw(T c d−1 ) ≥ 2, then undo the partial routing that we did earlier; instead we go eastward from d−and then upward to d L = d R in row 1.If d = s + 2 and r pw(T c d−1 ) = 1, then the partial drawing of

Figure 9 :
Figure 9: The constructions if the maximum degree is 3.

Figure 10 :Lemma 4 .
Figure 10: A Halin-graph requiring much more height than its pathwidth.(a) Tree C 1 with cycle C (cycle-edges are dotted red) that encloses c.(b) Obtaining F w from C w .Dashes edges are not needed except to avoid degree-2 vertices in the trees.(c) Obtaining C w+1 using many copies of F w .(d) Tree Ĉ2 needed for Theorem 6.

Figure 12 :
Figure 12: (a) Three bad non-spine children of type (t, b) imply a planar drawing of K 3,3 .(Picture based on [1]).(b) Possible arrangements of non-spine children of s j 2 on layer + 1.

Figure 13 :
Figure 13: For the proof of Claim 6.(a) Poly-line Q separates I from C. (b) The pocket P .