Finding Tutte paths in linear time

TL;DR

This paper presents a new, simpler proof and a linear-time algorithm for finding Tutte paths in 3-connected planar graphs, with applications to spanning trees and walks.

Contribution

It provides a new proof and a linear-time algorithm for Tutte paths in 3-connected planar graphs, with enhanced properties and applications.

Findings

Linear-time algorithm for Tutte paths in 3-connected planar graphs

Tutte paths visit all exterior vertices and have exactly three attachment points per component

Applications include binary spanning trees and 2-walks

Abstract

It is well-known that every planar graph has a Tutte path, i.e., a path $P$ such that any component of $G-P$ has at most three attachment points on $P$. However, it was only recently shown that such Tutte paths can be found in polynomial time. In this paper, we give a new proof that 3-connected planar graphs have Tutte paths, which leads to a linear-time algorithm to find Tutte paths. Furthermore, our Tutte path has special properties: it visits all exterior vertices, all components of $G-P$ have exactly three attachment points, and we can assign distinct representatives to them that are interior vertices. Finally, our running time bound is slightly stronger; we can bound it in terms of the degrees of the faces that are incident to $P$. This allows us to find some applications of Tutte paths (such as binary spanning trees and 2-walks) in linear time as well.

Figures

Equations

$$(⊐)$$

Full text

1 Introduction

A Tutte path is a well-known generalization of Hamiltonian paths that allows to visit only a subset of the vertices of the graph, as long as all remaining vertices are in components with at most three attachment points. (Detailed definitions are below.) They have been studied extensively, especially for planar graphs, starting from Tutte’s original result:

Theorem 1** ([24]).**

Let $G$ be a 2-connected planar graph with distinct vertices $X,Y$ on the outer face. Let $\alpha$ be an edge on the outer face. Then $G$ has a Tutte path from $X$ to $Y$ that uses edge $\alpha$.

We refer to the recent work by Schmid and Schmidt [20] for a detailed review of the history and applications of Tutte paths. It was long not known how to compute a Tutte path in less than exponential time. A breakthrough was achieved by Schmid and Schmidt in 2015 [18, 19], when they showed that one can find a Tutte path for 3-connected planar graphs in polynomial time. In 2018, the same authors then argued that Tutte paths can be found in polynomial time even for 2-connected planar graphs [20]. For both papers, the main insight is to prove the existence of a Tutte path by splitting the graph into non-overlapping subgraphs to recurse on; the split can be found in linear time and therefore the running time becomes quadratic.

In this paper, we show that Tutte paths can be computed in linear time. To do so, we give an entirely different proof of the existence of a Tutte path for 3-connected planar graph. This proof is very simple if the graph is triangulated, (we give a quick sketch below), but requires more care when faces have larger degrees. Our path (and also the one by Schmid and Schmidt [18, 19]) comes with a system of distinct representatives, i.e., an injective assignment from the components of $G\setminus P$ to vertices of $P$ that are attachment points. Such representatives are useful for various applications of Tutte paths.

Our proof for 3-connected planar graphs is based on a Hamiltonian-path proof by Asano, Kikuchi and Saito [1] that was designed to give a linear-time algorithm; with arguments much as in their paper we can therefore find the Tutte path and its representatives in linear time. Since 3-connected planar graphs are (as we argue) the bottleneck in finding Tutte paths, this shows that the path of Theorem 1 can be found in linear time.

1.1 Preliminaries

We assume familiarity with graphs, see, e.g., Diestel [8]. Throughout this paper, $G=(V,E)$ denotes a graph with $n$ vertices and $m$ edges. We assume that $G$ is planar, i.e., can be drawn in 2D without edge crossings. A planar drawing of $G$ splits $\mathbb{R}^{2}$ into connected regions called faces; the unbounded region is the outer face while all others are called interior faces. A vertex/edge is called exterior if it is incident to the outer face and interior otherwise. We assume throughout that $G$ is plane, i.e., one particular abstract drawing of $G$ has been fixed (by giving the clockwise order of edges around each vertex and the edges that are on the outer face). Any subgraph of $G$ inherits this planar embedding, i.e., uses the induced order of edges and as outer face the face that contained the outer face of $G$. The following notion will be convenient: Two vertices $v$ and $w$ are interior-face-adjacent (in a planar graph $G$) if there exists an interior face that is incident to both $v$ and $w$. We will simply write face-adjacent since we never consider adjacency via the outer face.

Nooses and connectivity. For a fixed planar drawing of $G$, let a noose be a simple closed curve $\mathcal{N}$ that goes through vertices and faces and crosses no edge except at endpoints. Note that a noose can be described as a cyclic sequence $\langle x_{0},f_{1},x_{1},\dots,f_{s},x_{s}{=}x_{0}\rangle$ of vertices and faces such that $f_{i}$ contains $x_{i-1}$ and $x_{i}$, and hence is independent of the chosen drawing. Frequently, the choice of faces will be clear from context or irrelevant; we then say that ${\mathcal{N}}=\langle x_{0},\dots,x_{s}{=}x_{0}\rangle$ goes through $\{x_{1},\dots,x_{s}\}$. The subgraph inside/outside $\mathcal{N}$ is the graph induced by the vertices that are on or inside/outside $\mathcal{N}$. The subgraph strictly inside/outside is obtained from this by deleting the vertices on $\mathcal{N}$.

A graph $G$ is connected if for any two vertices $v,w$ there is a path from $v$ to $w$ in $G$. A cutting $k$-set in $G$ is a set $S=\{x_{1},\dots,x_{k}\}$ of vertices such that $G\setminus S$ has more connected components than $G$. We call it a cutting pair for $k=2$ and a cutting triplet for $k=3$. A graph $G$ is called $k$-connected if it has no cutting $(k-1)$-set. Since we are only studying planar graphs, it will be convenient to use a characterization of connectivity via nooses. Consider a noose $\mathcal{N}$ that goes through $\{x_{1},\dots,x_{k}\}$ (and no other vertices), and there are vertices both strictly inside and strictly outside $\mathcal{N}$. Then clearly $S=\{x_{1},\dots,x_{k}\}$ is a cutting $k$-set. Vice versa, it is not hard to see that in a planar graph, any cutting $k$-set $S$ for $k=1,2,3$ gives rise to a noose $\mathcal{N}$ through $S$ that has vertices both strictly inside and strictly outside. A cut component $C$ of $S$ is a subgraph strictly inside a noose $\mathcal{N}$ through some of the vertices of $S$ such that $C$ contains at least one vertex not in $S$ and is inclusion-minimal among all such nooses. In particular, a cut component $C$ contains no vertices or edges of $S$, but in our algorithm we will frequently add the vertices and edges of $S$ to $C$ and call the result $C^{+}$.

Hamiltonian paths and Tutte paths. A Hamiltonian path is a path that visits every vertex exactly once. To generalize it to Tutte paths, we need more definitions. Fix a path $P$ in the graph. A $P$-bridge $C$ is a cut component of $P$; its attachment points its vertices on $P$.111Our definition of $P$-bridge considers only the proper $P$-bridges [24] that contain at least one vertex. A Tutte path is a path $P$ such that any $P$-bridge $C$ has at most three attachment points, and if $C$ contains exterior edges, then it has at most two attachment points. Our Tutte paths for 3-connected graphs will be such that no $P$-bridges contain exterior edges, so the second restriction holds automatically.

A Tutte path with a system of distinct representatives (SDR), also called a ${T_{\mathrm{SDR}}}$-path for short, is a Tutte path $P$ together with an injective assignment $\sigma$ from the $P$-bridges to vertices in $P$ such that for every $P$-bridge $C$ vertex $\sigma(C)$ is an attachment point of $C$.

Given a path $P$ in a plane graph, we denote by $F(P)$ the set of all interior faces that contain at least one vertex of $P$.

1.2 From 3-connected to 2-connected

In this section, we show that, to find the path of Theorem 1 efficiently, it suffices to consider 3-connected planar graphs.

We re-prove Theorem 1, presuming it holds for 3-connected planar graphs, by induction on the number of vertices with an inner induction on the number of exterior vertices. Say we want to find a Tutte path from $X$ to $Y$ that uses exterior edge $\alpha=(U,W)$, where $X,Y$ are exterior vertices. In the base case, $G$ is 3-connected and we are done. So assume that $G$ has cutting pairs. If edge $(X,Y)$ does not exist, then add it in such a way that $\alpha$ stays exterior, and find a Tutte path $P$ in the resulting graph recursively (it has fewer exterior vertices). Since $\{X,Y\}\neq\{U,W\}$ (because $(U,W)\in G$ while $(X,Y)\not\in G$), path $P$ visits at least one vertex other than $X,Y$, and so cannot use edge $(X,Y)$. So it is also a Tutte path of $G$.

Now, assume that $(X,Y)$ exists. Repeatedly split the graph at any cutting pair $\{u,v\}$ into cut components $C_{1},\ldots,C_{k}$, and store the 3-connected components $C_{1}^{+},\ldots,C_{k}^{+}$—induced by the cut components and $u,v$ and inserting a virtual edge $(u,v)$—in a so-called SPQR-tree [7, 12], which additionally creates one leaf node for every edge of $G$. This can be done in linear time [13].

Root the SPQR-tree at the node of edge $(X,Y)$. For each 3-connected component $C^{+}$ other than the root, set $\{X_{C},Y_{C}\}$ to be the cutting pair that $C^{+}$ has in common with its parent component, and observe that these two vertices are necessarily exterior in $C$ since $X,Y$ are exterior in $G$; see Fig. 1.

If $C^{+}$ has only these two vertices, then let $P_{C}$ be the path $(X_{C},Y_{C})$. Otherwise, define an edge $\alpha_{C}\neq(X_{C},Y_{C})$ of $C^{+}$ as follows: If the node of $\alpha$ is a descendant of $C^{+}$, then let $\alpha_{C}$ be the virtual edge of $C^{+}$ that it shares with the child that leads to this descendant. Note that $\alpha_{C}$ is a virtual edge, and it is necessarily on the outer face of $C$ since $\alpha$ is on the outer face of $G$. Otherwise ($\alpha$ is not in a descendant of $C^{+}$) choose $\alpha_{C}$ to be an arbitrary exterior edge of $C$ other than $(X_{C},Y_{C})$. Let $P_{C}$ be a Tutte path that begins at $X_{C}$, ends at $Y_{C}$ and uses edge $\alpha_{C}$; we know that this exists since $C^{+}$ is either a triangle or a 3-connected graph.

Now, obtain the Tutte path $P$ of $G$ by repeatedly substituting paths of 3-connected components. Specifically, initiate $P$ as the virtual copy of edge $(X,Y)$ that was added when we created the node for $(X,Y)$. For as long as $P$ contains a virtual edge $(u,v)$, let $C^{+}$ be the child component at this virtual edge and observe that $\{X_{C},Y_{C}\}=\{u,v\}$. Substitute $P_{C}$ in place of edge $(u,v)$ of $P$, i.e., set $P$ to be X\mathrel{\leavevmode\hbox to10.06pt{\vbox to12.04pt{\pgfpicture\makeatletter\hbox{\hskip 5.02759pt\lower-7.41957pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{ }\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ } {{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{ {}{}}}{ {}{}} {{}{{}}}{{}{}}{}{{}{}} { }{{{{}}\pgfsys@beginscope\pgfsys@invoke{ }\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{-2.73315pt}{-2.72516pt}\pgfsys@invoke{ }\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{ }\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{\vbox{\halign{\hfil#\hfil\cr\cr\vskip 0.667pt\cr\hbox{{\scriptstyle P}}\cr}}}} }}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}} { {}}{}{{}}{}{{}} {}{} {}{{}{}}{}{}{}{{}}{{}}{{}{}}{{}{}} {{{{}{}{{}} }}{{}} }{{{{}{}{{}} }}{{}}}{{{{}{}{{}} }}{{}}{{}}} {{{{}{}{{}} }}{{}} {} }{{{{}{}{{}} }}{{}} {} }{{{{}{}{{}} }}{{}} {} }{{{{}{}{{}} }}{{}} {{}} } {{{{}{}{{}} }}{{}}{{}}} {}{{ {\pgfsys@beginscope \pgfsys@setdash{}{0.0pt}\pgfsys@roundcap\pgfsys@roundjoin{} {}{}{} {}{}{} \pgfsys@moveto{-2.07999pt}{2.39998pt}\pgfsys@curveto{-1.69998pt}{0.95998pt}{-0.85318pt}{0.28pt}{0.0pt}{0.0pt}\pgfsys@curveto{-0.85318pt}{-0.28pt}{-1.69998pt}{-0.95998pt}{-2.07999pt}{-2.39998pt}\pgfsys@stroke\pgfsys@endscope}} }{}{}{{}}\pgfsys@moveto{4.4276pt}{-4.8196pt}\pgfsys@lineto{1.32759pt}{-4.8196pt}\pgfsys@lineto{0.47401pt}{-5.51959pt}\pgfsys@lineto{-1.23315pt}{-4.1196pt}\pgfsys@lineto{-2.94032pt}{-5.51959pt}\pgfsys@lineto{-4.82759pt}{-4.8196pt}\pgfsys@lineto{-4.82759pt}{-4.8196pt}\pgfsys@stroke\pgfsys@invoke{ }{{}{{}}{}{}{{}}{{{}}{{{}}{\pgfsys@beginscope\pgfsys@invoke{ }\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{4.6276pt}{-4.8196pt}\pgfsys@invoke{ }\pgfsys@invoke{ \lxSVG@closescope }\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}{{}}}} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{{ {}{}{}{}{}}}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}u/v\mathrel{\leavevmode\hbox to13.2pt{\vbox to13.41pt{\pgfpicture\makeatletter\hbox{\hskip 6.60008pt\lower-8.1029pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{ }\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ } {{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{ {}{}}}{ {}{}} {{}{{}}}{{}{}}{}{{}{}} { }{{{{}}\pgfsys@beginscope\pgfsys@invoke{ }\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{-4.30565pt}{-3.4085pt}\pgfsys@invoke{ }\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{ }\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{\vbox{\halign{\hfil#\hfil\cr\cr\vskip 0.667pt\cr\hbox{{\scriptstyle P_{C}}}\cr}}}} }}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}} { {}}{}{{}}{}{{}} {}{} {}{{}{}}{}{}{}{{}}{{}}{{}{}}{{}{}} {{{{}{}{{}} }}{{}} }{{{{}{}{{}} }}{{}}}{{{{}{}{{}} }}{{}}{{}}} {{{{}{}{{}} }}{{}} {} }{{{{}{}{{}} }}{{}} {} }{{{{}{}{{}} }}{{}} {} }{{{{}{}{{}} }}{{}} {} }{{{{}{}{{}} }}{{}} {} }{{{{}{}{{}} }}{{}} {{}} } {{{{}{}{{}} }}{{}}{{}}} {}{}{}{}{{}}\pgfsys@moveto{6.00009pt}{-5.50293pt}\pgfsys@lineto{2.90009pt}{-5.50293pt}\pgfsys@lineto{2.04651pt}{-6.20293pt}\pgfsys@lineto{0.33934pt}{-4.80293pt}\pgfsys@lineto{-1.36783pt}{-6.20293pt}\pgfsys@lineto{-3.075pt}{-4.80293pt}\pgfsys@lineto{-4.78217pt}{-6.20293pt}\pgfsys@lineto{-6.40009pt}{-5.50293pt}\pgfsys@lineto{-6.40009pt}{-5.50293pt}\pgfsys@stroke\pgfsys@invoke{ }{{}{{}}{}{}{{}}{{{}}{{{}}{\pgfsys@beginscope\pgfsys@invoke{ }\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{6.20009pt}{-5.50293pt}\pgfsys@invoke{ }\pgfsys@invoke{ \lxSVG@closescope }\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}{{}}}} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{{ {}{}{}{}{}}}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}v/u\mathrel{\leavevmode\hbox to10.06pt{\vbox to12.04pt{\pgfpicture\makeatletter\hbox{\hskip 5.02759pt\lower-7.41957pt\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ }\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{ }\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ } {{}}\hbox{\hbox{{\pgfsys@beginscope\pgfsys@invoke{ }{{}{}{{ {}{}}}{ {}{}} {{}{{}}}{{}{}}{}{{}{}} { }{{{{}}\pgfsys@beginscope\pgfsys@invoke{ }\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{-2.73315pt}{-2.72516pt}\pgfsys@invoke{ }\hbox{{\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{ }\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{ }\hbox{{\vbox{\halign{\hfil#\hfil\cr\cr\vskip 0.667pt\cr\hbox{{\scriptstyle P}}\cr}}}} }}\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}} { {}}{}{{}}{}{{}} {}{} {}{{}{}}{}{}{}{{}}{{}}{{}{}}{{}{}} {{{{}{}{{}} }}{{}} }{{{{}{}{{}} }}{{}}}{{{{}{}{{}} }}{{}}{{}}} {{{{}{}{{}} }}{{}} {} }{{{{}{}{{}} }}{{}} {} }{{{{}{}{{}} }}{{}} {} }{{{{}{}{{}} }}{{}} {{}} } {{{{}{}{{}} }}{{}}{{}}} {}{}{}{}{{}}\pgfsys@moveto{4.4276pt}{-4.8196pt}\pgfsys@lineto{1.32759pt}{-4.8196pt}\pgfsys@lineto{0.47401pt}{-5.51959pt}\pgfsys@lineto{-1.23315pt}{-4.1196pt}\pgfsys@lineto{-2.94032pt}{-5.51959pt}\pgfsys@lineto{-4.82759pt}{-4.8196pt}\pgfsys@lineto{-4.82759pt}{-4.8196pt}\pgfsys@stroke\pgfsys@invoke{ }{{}{{}}{}{}{{}}{{{}}{{{}}{\pgfsys@beginscope\pgfsys@invoke{ }\pgfsys@transformcm{1.0}{0.0}{0.0}{1.0}{4.6276pt}{-4.8196pt}\pgfsys@invoke{ }\pgfsys@invoke{ \lxSVG@closescope }\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope}}{{}}}} \pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope{{ {}{}{}{}{}}}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope }\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}}Y. Note that, if $C^{+}$ is not the singleton-edge $(u,v)$, then $P_{C}$ contains $\alpha_{C}$, which is a virtual edge. This means that the process repeats until we have substituted the real edges from the leaves of the SPQR-trees. In particular (due to our choice of $\alpha_{C}$), we will substitute the paths from all components between $(X,Y)$ and $\alpha$, which means that $\alpha$ is an edge of the final path $P$ as required.

Observe that for some 3-connected components we do not substitute their paths; these become $P$-bridges with two attachment points. There may also be some $P$-bridges within each 3-connected components, but these have at most three attachment points since we used a Tutte path for each component. So the result is the desired Tutte path. Since we compute one Tutte path per 3-connected component, and this can be done in time proportional to the size of the component, the overall running time is linear.

1.3 Simple proof for triangulated planar graphs

So it suffices to find Tutte paths in linear time for 3-connected planar graphs. As a convenient warm-up, we sketch here first the (much simpler) case of a triangulated planar graph $G$, i.e., every face is a triangle. Remove, for every non-facial triangle $T$, the graph in its interior. What remains is a 4-connected triangulated planar graph $H$, say it has $k$ vertices. The graph $H$ has a Hamiltonian path $P$, and Asano, Kikuchi, and Saito [1] showed how to find it in linear time. Studying their proof, one can easily verify that we can force that $P$ begins at $X$, ends at $Y$, and contains edge $\alpha$, for given $X,Y,\alpha$ on the outer face.222For readers familiar with [1]: It suffices to choose $A,B,C$ in the outermost recursion such that $A=X$, $B=Y$ and $C$ is the end of $\alpha$ that is closer to $B$. Note that $P$ automatically is a Tutte path, because every $P$-bridge resides inside an interior face of $H$, and hence has three attachment points and no edge on the outer face.

To find a system of representatives, we initially allow edges of $P$ to be representatives, but forbid using exterior vertices or the edge $\alpha$. Thus, assign to every interior face $T$ of $H$ a representative $\sigma(T)\in V(P)\cup E(P)$ such that $\sigma(T)$ is incident to $T$, not an exterior vertex or $\alpha$, and no two triangles obtain the same representative. There are $2k-5$ faces of $H$, and path $P$ has $k-2$ edges $\neq\alpha$ and $k-3$ interior vertices, so there are sufficiently many possible representatives. One can argue that for the Hamiltonian path from [1], it is possible to assign the faces to these representatives in an injective manner (we omit the details; they are implicit in our proof below.)

To obtain a ${T_{\mathrm{SDR}}}$-path, we apply a substitution trick (explained in detail below) for every edge $e\in P$ that is used as representative $\sigma(T)$ for some interior face $T$. Namely, if $T$ has no $P$-bridge inside, then simply remove $e$ as representative. Otherwise, replace $e$ by a (recursively obtained) ${T_{\mathrm{SDR}}}$-path $P_{T}$ of the $P$-bridge inside $T$. Since we can specify the ends of $P_{T}$, and it uses no vertices of its outer face $T$ as representatives, this gives a ${T_{\mathrm{SDR}}}$-path of $G$ after repeating at all edge-representatives. The overall running time is linear, because we can find $H$ by computing the tree of 4-connected components in linear time [14], and then compute a Hamiltonian path in each 4-connected component in linear time [1]. The ${T_{\mathrm{SDR}}}$-path is obtained by substituting the Hamiltonian paths of child-components into the one of the root as needed to replace edge-representatives.

2 Tutte paths in 3-connected planar graphs

For triangulated planar graphs, one can quite easily find a ${T_{\mathrm{SDR}}}$-path by removing the interiors of all separating triangles, and finding for the resulting 4-connected planar graph a Hamiltonian path using the approach of Asano, Kikuchi, and Saito [1]. It is not hard to see that we can assign representatives to all separating triangles, possibly after expanding the path using the substitution trick described below. (We omit the details for space reasons.)

For 3-connected planar graphs that are not triangulated, we use the same approach, but must generalize many definitions from [1] and add quite a few cases because now face-adjacent vertices are not necessarily adjacent. To keep the proof self-contained, we re-phrase everything from scratch.333Indeed, due to attempts to simplify the notations similar as done in [4], the reader familiar with [1] may barely see the correspondence between the proof and [1]. Roughly, their Condition (W) corresponds to $c3c(X,W,Y)$, their Case 1 is our Case 2.1.3, and their Case 3 combines our Case 2.1.2 with our Case 2.1.5 (but resolves it in a symmetric fashion).

We need a few definitions. The outer stellation of a planar graph $G$ is the graph obtained by adding a vertex in the outer face and connecting it to all exterior vertices. A planar graph $G$ is called internally 3-connected if its outer stellation is 3-connected. Note that this implies that $G$ is 2-connected, any cutting pair is exterior (i.e., has both vertices on the outer face) and has only two cut components that contain other vertices. In the following, we endow $G$ with $k$ corners, which are $k$ vertices $X_{1},\dots,X_{k}$ that appear in this order on the outer face. Usually, $k=3$ or $4$, but occasionally we allow larger $k$. A side of such a graph is the outer face path between two consecutive corners that does not contain any other corners. The corner stellation $G^{\mathrm{s}}$ is obtained by adding a vertex in the outer face and connecting it to the corners. We say that $G$ is corner-3-connected with respect to corners $X_{1},\dots,X_{k}$ (abbreviated to “$G$* satisfies $c3c(X_{1},\dots,X_{k})$*”) if $G^{\mathrm{s}}$ is 3-connected. Figure 2(a) illustrates this condition. It is easy to show that $G$ satisfies $c3c(X_{1},\dots,X_{k})$ if and only if $k\geq 3$, $G$ is internally 3-connected, and no cutting pair $\{v,w\}$ of $G$ has both $v$ and $w$ on one side of $G$.

For ease of proof we make the induction hypothesis stronger than just having a ${T_{\mathrm{SDR}}}$-path, by restricting which vertices must be visited and which vertices must not be representatives. A ${T_{\mathrm{int}}}$-path is a ${T_{\mathrm{SDR}}}$-path $P$ that visits all exterior vertices, and where representative $\sigma(C)$ is interior, for all $P$-bridges $C$. The goal of the remainder of this section is to prove the following result (which immediately implies Theorem 1 for 3-connected graphs444Theorem 1 allows $(U,W)=(X,Y)$, but then holds trivially since using edge $(X,Y)$ as path satisfies all conditions. We require $(U,W)\neq(X,Y)$ since we want not just a Tutte path but a ${T_{\mathrm{int}}}$-path, and the single-edge path $(X,Y)$ would allow only exterior vertices as representatives.):

Lemma 1**.**

Let $G$ be a plane graph with distinct vertices $X,Y$ on the outer face. Let $(U,W)\neq(X,Y)$ be an edge on the outer face. If $G$ satisfies $c3c(X,U,W,Y)$, then it has a ${T_{\mathrm{int}}}$-path that begins at $X$, ends at $Y$, and contains $(U,W)$.

We need a second result for the induction. Let a ${T_{\mathrm{end}}}$-path be a ${T_{\mathrm{SDR}}}$-path $P$ that visits all exterior vertices, and where representative $\sigma(C)$ is interior or the last vertex of $P$, for all $P$-bridges $C$.

Lemma 2**.**

Let $G$ be a plane graph with distinct vertices $X,Y$ on the outer face. Let $(U,W)\neq(X,Y)$ be an edge on the outer face. If $G$ satisfies $c3c(X,U,W,Y)$ and

[TABLE]

then $G$ has a $T_{\mathrm{end}}$-path $P$ that begins at $X$, ends at $Y$, and uses $(U,W)$ and $(W,Y)$.555This lemma is a special case of the “Three Edge Lemma” [22], which states that for any three edges on the outer face there exists a Tutte cycle containing them all. However, it cannot simply be obtained from it since we require restrictions on the location of representatives. Further, if $Y$ is the representative of a $P$-bridge $C$, then $C$ has $W$ and $Y$ as attachment points.

See Figure 5(c) for a graph that satisfies 2.

We assume throughout that $X,U,W,Y$ are enumerated in ccw order along the outer face, the other case can be resolved by reversing the planar embedding.

The following trick will help shorten the proof: If graph $G$ satisfies 2, then Lemma 2 implies Lemma 1. Namely, if Lemma 2 holds then we have a $T_{\mathrm{end}}$-path $P$ from $X$ to $Y$ through $(U,W)$ and $(W,Y)$. If this is not a $T_{\mathrm{int}}$-path, then some $P$-bridge $C$ has $Y$ as representative, and by assumption also has $W$ as attachment point. It must have a third attachment point $u$, otherwise $\{W,Y\}$ would be a cutting pair within one side of $G$, contradicting corner-3-connectivity. It has no more attachment points since $P$ is a Tutte path, so $\{W,Y,u\}$ is a cutting triplet. We apply the substitution trick described below (and useful in other situations as well), which replaces $(W,Y)$ with a path through $C$ that does not use $u$. Thus, $C$ no longer needs a representative and we obtain a $T_{\mathrm{int}}$-path.

The substitution trick. This trick can be applied whenever we have an edge $e=(w,y)$ used by some ${T_{\mathrm{SDR}}}$-path $P$, and a $P$-bridge $C$ that resides inside a noose through some cutting triplet $\{u,w,y\}$ for some vertex $u$. Define $C^{+}=G[C]\cup\{(u,w),(w,y)\}\setminus\{(u,y)\}$, where edges are added only if they did not exist in $G[C]$.666We apply the substitution trick even when $V(C^{+})=V(G)$ and $G$ has a triangular outer face; not adding edge $(w,y)$ will ensure that $C^{+}$ has fewer interior vertices and induction can be applied.

One easily verifies that $C^{+}$ satisfies $c3c(u,v,w)$, else there would have been a cutting pair in $G$. Hence, by induction, $C^{+}$ has a $T_{\mathrm{int}}$-path $P_{C^{+}}$ from $u$ to $y$ that uses edge $(u,w)$. It does not use the edge $(u,y)$ since $P_{C^{+}}$ begins at $u$ with edge $(u,w)$. So $P_{C^{+}}\setminus(u,w)$ is a path in $C$ from $w$ to $y$ that does not visit $u$. Substitute this in place of edge $(w,y)$ of $P$; see Fig. 2(b). One easily verifies that the resulting path $P^{\prime}$ is a $T_{\mathrm{int}}$-path. We will prove this in full detail in Section 4, but roughly speaking, combining paths preserves $T_{\mathrm{int}}$-paths because every $P^{\prime}$-bridge can inherit its representative from $P$ or $P_{C^{+}}$, and no vertex is used twice as representative since $P_{C^{+}}$ does not use $\{u,v,w\}$ as representatives.

2.1 Proof of Lemma 1 and Lemma 2

We prove the two lemmas simultaneous by induction on the number of vertices of $G$, with an inner induction on the number of interior vertices. The base case is $n=3$ where $G$ is a triangle, but the same construction works whenever the outer face is a triangle (see below). For the induction step, we need the notation $S_{xy}$, which is the outer face path from $x$ to $y$ in ccw direction. In particular, the four sides are $S_{XU}$, $S_{UW}$, $S_{WY}$ and $S_{YX}$. We sometimes name sides as suggested by Fig. 2(a), so $S_{XU},S_{UW},S_{WY}$ and $S_{YX}$ are the left/bottom/right/top side, respectively.

2.1.1 Case 1: The outer face is a triangle

Figure 2(c) illustrates this case. We know that $X\neq Y$ and $U\neq W$, so we must have $X=U$ or $W=Y$. For Lemma 2, we know that 2 holds, which forces $W\neq Y$, hence $X=U$. For Lemma 1, we may assume $X=U$ by symmetry, for otherwise we reverse the planar embedding, find a path from $Y$ to $X$ that uses $(W,U)$ (with this, we have $X^{\prime}=U^{\prime}$) and then reverse the result.

So $X=U$. Define $P$ to be $\langle X{=}U,W,Y\rangle$ and observe that this is a $T_{\mathrm{end}}$-path, because the unique $P$-bridge $C$ (if any) has attachment points $\{U,W,Y\}$, and we can assign $Y$ to be its representative. So Lemma 2 holds. Since condition 2 is satisfied, this implies Lemma 1.

2.1.2 Case 2: $G$ has a cutting pair $\{u,w\}$

with $u$ and $w$ on the left and right side

Figure 3 illustrates this case. Let ${\mathcal{N}}$ be a noose through $u$ and $w$ along a common interior face $f^{*}$ and then going through the outer face. Let $G_{t}$ and $G_{b}$ be the subgraphs inside and outside ${\mathcal{N}}$, named such that $G_{b}$ contains the bottom side. Let $G_{t}^{+}/G_{b}^{+}$ be the graphs obtained from $G_{t}/G_{b}$ by adding $(u,w)$ to each, even if it did not exist in $G$ (we will ensure that the final path does not use it).

We first show Lemma 1. One can easily verify that $G_{t}$ satisfies $c3c(X,u,w,Y)$ since its outer face is a simple cycle; see Appendix 4. Apply induction and find a $T_{\mathrm{int}}$-path $P_{t}$ of $G^{+}_{t}$ from $X$ to $Y$ that uses edge $(u,w)$. Now apply a modified substitution trick to $(u,w)$. Namely, by induction, there is a $T_{\mathrm{int}}$-path $P_{b}$ of $G^{+}_{b}$ from $u$ to $w$ that uses edge $(U,W)$. Substitute $P_{b}$ into $P_{t}$ in place of $(u,w)$ to get $P$. Path $P$ uses $(U,W)$ since $P_{b}$ does. It does not use $(u,w)$ since we removed this from $P_{t}$, and $P_{b}$ cannot use it since $P_{b}$ starts at $u$, ends at $w$, and visits $(U,W)$ in between. So after inheriting representatives from $P_{b}$ and $P_{t}$ we obtain a $T_{\mathrm{int}}$-path $P$ in $G$.

To prove Lemma 2, note that exactly one of $G^{+}_{t}$ and $G^{+}_{b}$ contains $(W,Y)$; use a $T_{\mathrm{end}}$-path for this subgraph and create $P$ as above. Only one graph uses $Y$ as representative, and one easily shows that $P$ is a $T_{\mathrm{end}}$-path.

2.1.3 Case 3: $G$ has a cutting pair $\{y,w\}$ with

$y$ and $w$ on the top and right side, respectively. Furthermore, there is an interior face $f^{*}$ containing $y$ and $w$ that does not contain $Y$.

For later applications, we first want to point out that if $G$ has a cutting pair $\{y,w\}$ on the top and right side for which $(y,w)$ is an edge, then such a face $f^{*}$ always exists, because there are two interior faces containing $y$ and $w$, and not both can contain $Y$.

Figure 4 illustrates this case. We know $w\neq Y\neq y$, else $\{y,w\}$ would be a cutting pair within one side. We may assume $y\neq X$; else we can use Case 2.1.2. Hence the top side contains at least three vertices $X,y,Y$, so 2 does not hold and we have to prove only Lemma 1.

We choose $\{y,w\}$ such that $w$ is as close to $W$ as possible (along the right side). The face $f^{*}$ containing $y,w$ may have multiple edges on the top side; let $y$ be the one that is as close to $Y$ as possible. Define $G_{b}$, $G_{b}^{+}$, $G_{t}$, $G_{t}^{+}$ to be as in Case 2.1.2. Since the outer face of $G^{+}_{b}$ is a simple cycle, it satisfies $c3c(X,U,W,w,y)$. But since we chose $w$ to be as close to $W$ as possible, it also satisfies $c3c(X,U,W,y)$. Namely, assume for contradiction that some cutting pair $\{y^{\prime},w^{\prime}\}$ exists along the side $S_{Ww}\cup(w,y)$ of $G_{b}^{+}$; see Fig. 4(a). Since there is no cutting pair within $S_{Ww}$, it must have the form $\{y,w^{\prime}\}$ for some $w^{\prime}\neq w$ on $S_{Ww}$. As $f^{*}$ does not contain $Y$, neither can any face containing $\{y,w^{\prime}\}$, so $\{y,w^{\prime}\}$ could have been used for Case 2.1.3, contradicting our choice of $w$.

By induction, we can find a $T_{\mathrm{int}}$-path $P_{b}$ of $G^{+}_{b}$ from $X$ to $y$ that includes the edge $(U,W)$. The plan is to combine $P_{b}$ with a path through $G_{t}$, but we must distinguish cases.

Case 3a: $P_{b}$ does not contain $(y,w)$ or $(y,w)\in G$. Observe that $G_{t}^{+}$ satisfies $c3c(y,w,Y)$. By induction, find a $T_{\mathrm{int}}$-path $P_{t}$ in $G_{t}^{+}$ from $Y$ to $w$ that uses edge $(y,w)$. Append the reverse of $P_{t}\setminus(y,w)$ to $P_{b}$ to obtain a $T_{\mathrm{int}}$-path; see Fig. 4(b).

Case 3b: $P_{b}$ contains $(y,w)$ and $(y,w)\not\in G$. In this case, we must remove $(y,w)$ from the path and hence use a subpath in $G_{t}$ to reach vertex $y$. This requires further subcases. Let $\pi_{f}$ be the path along $f^{*}$ from $y$ to $w$ that becomes part of the the outer face of $G_{t}$. Let $(y,z)$ be the edge incident to $y$ on $\pi_{f}$.

Case 3b-1: $\pi_{f}$ contains no vertex on the outer face of $G$ other than $y$ and $w$. See Fig. 4(c). The outer face of $G_{t}$ is then a simple cycle and $G_{t}$ satisfies $c3c(w,y,Y)$. By induction, we can find a ${T_{\mathrm{int}}}$-path $P_{t}$ in $G_{t}$ that begins at $Y$, ends at $w$, and uses $(y,z)$.

Case 3b-2: $\pi_{f}$ contains a vertex $x\neq y,w$ on the outer face of $G$. See Fig. 4(d). Since $x$ is on $f^{*}$, it cannot be on the top side by choice of $y$. So $x\in S_{wY}\setminus{Y}$. In fact, $x$ must be the neighbor of $w$ on both $S_{wY}$ and $\pi_{f}$, else there would be a cutting pair within the right side. Set $G_{t}^{\prime}$ to be the graph inside a noose through $y$ and $x$ that has $Y$ inside. Since $\pi_{f}$ has no vertices other than $y,x,w$ on the outer face of $G$, graph $G_{t}^{\prime}$ has a simple cycle as outer face, and therefore satisfies $c3c(Y,y,x)$, so it satisfies $c3c(Y,y,z,x)$. By induction, we can find a ${T_{\mathrm{int}}}$-path $P_{t}^{\prime}$ of $G_{t}$ that begins at $Y$, ends at $x$, and uses $(y,z)$. We append $(w,x)$ to obtain $P_{t}$.

In both cases, we obtain a path $P_{t}$ that begins at $Y$, ends at $w$, and visits all of $G_{t}$. Appending the reverse of this to $P_{b}\setminus(y,w)$ gives the $T_{\mathrm{int}}$-path.

2.1.4 Case 3*′*: $G$ has a cutting pair $\{y,w\}$ with

$y$ and $w$ on the top and left side, respectively. Furthermore, there is an interior face $f^{*}$ containing $y$ and $w$ that does not contain $X$.

This is handled symmetrically to Case 3.

2.1.5 Case 4: None of the above

In this case, we split $G$ into one big graph $G_{0}$ and (possibly many) smaller graphs $G_{1},\dots,G_{s}$, recurse in $G_{0}$ and then substitute $T_{\mathrm{int}}$-paths of $G_{1},\dots,G_{s}$ or use them as $P$-bridges.

We need two subcases, but first give some steps that are common to both. Let $Y_{X}$ be the neighbor of $Y$ on the top side. Define a $B$-necklace (for $B\in\{U,W\}$) to be a noose ${\mathcal{N}}_{0}:\langle Y_{X}{=}x_{0},f_{1},x_{1},\dots,x_{s-1},\allowbreak f_{s},x_{s}{=}B,f_{o}\rangle$, (where $f_{o}$ is the outer face) for which $x_{i}$ is face-adjacent to at least one vertex on $S_{WY}\setminus\{B\}$ for $1\leq i\leq s-1$. See also Fig. 5. We say that the necklace is simple if it contains no vertex twice, and interior if every $x_{i}$ (for $0<i<s$) is an interior vertex. One can argue that if none of the previous cases applies, then there always exists a simple interior $B$-necklace (see Section 4).

Route ${\mathcal{N}}_{0}$ through the outer face such that the left side is in its interior, and let $G_{0}$ (the “left graph”) be the graph inside ${\mathcal{N}}_{0}$. We say that ${\mathcal{N}}_{0}$ is leftmost if (among all simple interior $B$-necklaces) its left graph $G_{0}$ is smallest, and (among all simple interior $B$-necklaces whose left graph is $G_{0}$) it contains the most vertices of $G_{0}$. Fix a leftmost $B$-necklace $\langle x_{0},\dots,x_{s}\rangle$.

Claim 1**.**

If $(x_{i},x_{i+1})$ is not an edge for some $0\leq i<s$, then the face $f_{i}$ of ${\mathcal{N}}_{0}$ contains no vertex of $S_{WY}\setminus\{B\}$.

Proof.

If $(x_{i},x_{i+1})$ is not an edge of $G$, then both paths from $x_{i}$ to $x_{i+1}$ on $f_{i}$ contain at least one other vertex. One of them, say $z$, is inside ${\mathcal{N}}_{0}$. If $f_{i}$ contains a vertex of $S_{WY}\setminus\{B\}$, then $x_{i}$ and $z$ are face-adjacent, $z$ and $x_{i+1}$ are face-adjacent, and $z$ has a neighbor on $S_{WY}\setminus\{B\}$, so $x_{0},\ldots,x_{i},z,x_{i+1},\ldots,x_{s}$ is a simple interior $B$-necklace with the same left graph but containing more vertices of $G_{0}$. Hence ${\mathcal{N}}_{0}$ is not leftmost, a contradiction. ∎

For $i=0,\dots,{s-1}$, let $t_{i}$ be the vertex on $S_{WY}\setminus\{B\}$ that is face-adjacent to $x_{i}$ and closest to $Y$ (along the right side) among all such vertices. Set $t_{s}=W$ if $x_{s}=U$, and $t_{s}=Y$ otherwise. For $0<i\leq s$, define $\mathcal{N}_{i}$ to be the noose through $\langle x_{i-1},x_{i},t_{i},t_{i-1}\rangle$ such that the left side is outside ${\mathcal{N}}_{i}$. For $0<i\leq s$ let $G_{i}$ be the graph inside ${\mathcal{N}}_{i}$ (i.e., a cut component of $\{x_{i-1},x_{i},t_{i-1},t_{i}\}$); see Fig. 5(b).

Let $G^{+}$ be the graph obtained from $G$ by adding virtual edges $(x_{i},x_{i-1})$ and $(t_{i},t_{i-1})$ (for $i=1,\dots,s$) whenever these two vertices are distinct and the edge did not exist in $G$. Let $G_{0}^{+}$ be the graph obtained from $G_{0}$ by likewise adding virtual edges $(x_{0},x_{1}),\dots,(x_{s-1},x_{s})$. This makes the outer face of $G_{0}$ a simple cycle, so $G_{0}^{+}$ satisfies $c3c(X,U,B{=}x_{s},\dots,x_{0}{=}Y_{X})$. We distinguish two cases.

Case 4a: 2 holds, i.e., $(X,Y)$ and $(W,Y)$ are edges. We only have to prove Lemma 2 since this implies Lemma 1. Consider Figs. 5(c) and 5(d). Let $\langle x_{0}{=}Y_{X}{=}X,x_{1},\dots,x_{s}{=}W\rangle$ be a leftmost $W$-necklace. By $S_{WY}=(W,Y)$, we have $t_{i}=Y$ for all $i$. Since $x_{0}=Y_{X}=X$, we have that $G_{0}^{+}$ satisfies $c3c(X{=}x_{0},x_{1},\dots,x_{s}{=}W,U)$. But observe that $G_{0}^{+}$ has no cutting pair $\{x_{i},x_{j}\}$ with $0\leq i<j\leq s$, for otherwise the face $f$ containing $x_{i}$ and $x_{j}$ could be used as a shortcut and ${\mathcal{N}}_{0}$ was not leftmost (see Fig. 5(a)). So $G_{0}^{+}$ actually satisfies $c3c(X,W,U)$. Use induction to obtain a $T_{\mathrm{int}}$-path $P_{0}$ from $X$ to $W$ in $G_{0}^{+}$ that uses edge $(U,W)$. Then $P^{+}=P_{0}\cup(W,Y)$ is a path in $G^{+}$ that contains $(U,W)$, and $(W,Y)$.

Fix some $i=1,\dots,s$. If $P^{+}$ used edge $(x_{i-1},x_{i})$ and it was virtual, then by Claim 1 $f_{i}$ contains no vertex of $S_{WY}$, which means that the interior of $G_{i}$ is non-empty. Apply the substitution trick to remove $(x_{i-1},x_{i})$ from $P^{+}$, replacing it with a path through $G_{i}$. Otherwise, we keep $G_{i}$ as a $P^{+}$-bridge. We let its representative be $x_{i}$ if $1\leq i<s$, and $Y$ if $i=s$. Observe that this representative is interior or $Y$, and was not used by $P_{0}$ since $P_{0}$ was a $T_{\mathrm{int}}$-path. So we obtain a $T_{\mathrm{end}}$-path with the desired properties.

Case 4b: 2 does not hold. We must prove only Lemma 1 and may therefore by symmetry assume that $X\neq U$. We claim that this implies that $\deg(Y_{X})\geq 3$. For if $\deg(Y_{X})=2$, then its neighbors form a cutting pair, which by corner-3-connectivity means that $Y_{X}$ is a corner, hence $Y_{X}=X$. Since $X\neq U$, the two neighbors of $Y_{X}$ are then $Y$ and a vertex on the left side, and we could have applied Case 2.1.2. Contradiction, so $\deg(Y_{X})\geq 3$. Let $(Y_{X},x_{1})$ be the edge at $Y_{X}$ that comes after $(Y_{X},Y)$ in clockwise order (see Fig. 6(a)). Note that $x_{1}$ is face-adjacent to $Y$. It must be an interior vertex, for otherwise by $\deg(Y_{X})\geq 3$ edge $(Y_{X},x_{1})$ is a cutting pair that we could have used for Case 2.1.3 or 2.1.4.

Let ${\mathcal{N}}_{0}=\langle x_{0}{=}Y_{X},x_{1},\dots,x_{s}{=}U\rangle$ be a simple interior $U$-necklace; see Fig. 6(a). We use a $U$-necklace that is leftmost among all $U$-necklaces that contain $x_{1}$. Note that Claim 1 holds for ${\mathcal{N}}_{0}$ even with this restriction, since $(x_{0},x_{1})$ is an edge. We know that $G_{0}^{+}$ satisfies $c3c(X,Y_{X},x_{1},\dots,x_{s}{=}U)$. But observe that $G_{0}^{+}$ has no cutting pair $\{x_{i},x_{j}\}$ for $1\leq i<j\leq s$, for otherwise (as in Case 2.1.5) ${\mathcal{N}}_{0}$ would not be the leftmost necklace that uses $x_{1}$. So $G_{0}^{+}$ actually satisfies $c3c(X,Y_{X},x_{1},U)$.

By induction, obtain a $T_{\mathrm{int}}$-path $P_{0}$ in $G_{0}$ from $U$ to $X$ through edge $(x_{1},x_{0})$. Append the path $\langle U,W,t_{s},\ldots,t_{0}{=}Y\rangle$ to the reverse of $P_{0}$ to obtain path $P^{+}$. This path begins at $X$, ends at $Y$, and contains $(U,W)$. Any $P^{+}$-bridge is either a $P_{0}$-bridge (and receives a representative there) or is $G_{i}$ for some $1\leq i\leq s$. For $i>1$, assign representative $x_{i-1}$ to $G_{i}$; this is an interior vertex. Graph $G_{1}$ has an empty interior by choice of $x_{1}$, so it needs no representative.

There are two reasons why we cannot always use $P^{+}$ for the result. First, it may use virtual edges and hence not be a path in $G$. Second, some $P^{+}$-bridge $G_{i}$ may have four attachment points. Both are resolved by expanding $P^{+}$ via paths through the $G_{i}$’s. Fix one $i$ with $1\leq i\leq s$ and consider the following cases:

Case 4b-1: $(x_{i-1},x_{i})$ is virtual and used by $P^{+}$, and $t_{i-1}=t_{i}$. By Claim 1, the interior of graph $G_{i}$ is non-empty and inside the separating triplet $\{x_{i-1},x_{i},t_{i}\}$. Replace $(x_{i-1},x_{i})$ by a path through $G_{i}$ with the substitution trick, see graph $G_{3}$ in Fig. 6.

Case 4b-2: $(x_{i-2},x_{i})$ is virtual and used by $P^{+}$, and $t_{i-1}\neq t_{i}$. See Fig. 6(b)(top). We want to replace both $(x_{i-1},x_{i})$ and $(t_{i-1},t_{i})$ (which is always used by $P^{+}$) with a path through $G_{i}$. Let $G_{i}^{\asymp}$ be the graph $G_{i}$ with $(t_{i},x_{i})$ and $(t_{i-1},x_{i-1})$ added. The outer face of $G_{i}^{\asymp}$ is a simple cycle since $f_{i}$ contains no vertex of the right side by Claim 1, so $G_{i}^{\asymp}$ satisfies $c3c(t_{i},t_{i-1},x_{i-1},x_{i})$. By induction, find a $T_{\mathrm{int}}$-path $P_{i}$ in $G_{i}^{\asymp}$ from $t_{i}$ to $x_{i}$ that uses the edge $(t_{i-1},x_{i-1})$. So removing $(t_{i-1},x_{i-1})$ from $P_{i}$ splits it into two paths: path $P_{i}^{\operatorname{R}}$ connects $t_{i}$ to $t_{i-1}$, and path $P_{i}^{\operatorname{L}}$ connects $x_{i-1}$ to $x_{i}$. (No other split is possible by planarity.) Neither path uses the added edge $(t_{i},x_{i})$ since it connects the ends of $P_{i}$. Use $P_{i}^{\operatorname{R}}$ to replace $(t_{i-1},t_{i})$ and $P_{i}^{\operatorname{L}}$ to replace $(x_{i-1},x_{i})$ in $P^{+}$.

Case 4b-3: Subgraph $G_{i}$ has a non-empty interior and $t_{i}\neq t_{i-1}$. See Fig. 6(b)(bottom). In this case, $G_{i}$ is a $P^{+}$-bridge with four attachment points, a violation of Tutte path properties. If Case 2.1.5 applied to $G_{i}$, then $G_{i}$ is no longer a bridge of the resulting path and we are done. Otherwise, we do a substitution that uses a different supergraph of $G_{i}$.

Let $G_{i}^{\sqsubset}$ be $G_{i}$ with edges from path $\langle t_{i-1},x_{i-1},x_{i},t_{i}\rangle$ added if not already in $G_{i}$. This graph satisfies $c3c(t_{i},t_{i-1},x_{i-1},x_{i})$ and satisfies condition 2 if we set $X^{\prime}=t_{i}$, $U^{\prime}=t_{i-1},W^{\prime}=x_{i-1}$ and $Y^{\prime}=x_{i}$. So we can find a $T_{\mathrm{end}}$-path $P_{i}^{\prime}$ of $G_{i}^{\sqsubset}$ from $t_{i}$ to $x_{i}$ that uses $(t_{i-1},x_{i-1})$ and $(x_{i-1},x_{i})$. Thus, $P_{i}^{\prime}$ ends with $\langle t_{i-1},x_{i-1},x_{i}\rangle$ and $P_{i}^{\prime}\setminus\{(t_{i-1},x_{i-1}),(x_{i-1},x_{i})\}$ is a path from $t_{i-1}$ to $t_{i}$ in $G_{i}$ that does not visit $x_{i-1}$ or $x_{i}$. Substitute this path in place of edge $(t_{i-1},t_{i})$ in $P^{+}$. Note that one $P_{i}^{\prime}$-bridge $C$ may use $x_{i}$ as its representative, but if so, then it also has $x_{i-1}$ as attachment point. We set $x_{i-1}$ (which was $G_{i}$’s representative and is no longer needed as such) to be the representative of $C$.

Case 4b-4: $t_{i-1}\neq t_{i}$ and $(t_{i-1},t_{i})$ is virtual. Since $P^{+}$ always uses edge $(t_{i-1},t_{i})$, we must replace this edge with a path through $G_{i}$. We claim that this is done automatically because Case 2.1.5 applies. Namely, if $(t_{i-1},t_{i})$ is virtual, then there is at least one vertex between $t_{i-1}$ and $t_{i}$ on the right side. This vertex is exterior in $G$ and hence neither $x_{i}$ nor $x_{i-1}$. So it is strictly inside ${\mathcal{N}}_{i}$, hence $G_{i}$ has a non-empty interior and (by $t_{i-1}\neq t_{i}$) Case 4b-3 applies.

After doing these substitutions, there are no virtual edges in the path, no bridges have four attachment points, every bridge has an interior vertex as representative, and no vertex was used twice as representative; see Fig. 6(c). This ends the proof of Lemma 1 and 2.

2.2 Linear time complexity

It should be clear that our proof is algorithmic. The main bottlenecks for its running time are to determine which case to apply (i.e., whether there is a cutting pair) and to find the $B$-necklace. Both can be done in linear time, by computing all cutting pairs [7, 12] and by finding a leftmost path in the subgraph induced by vertices that are face-adjacent to $S_{WY}\setminus B$. This would yield quadratic running time overall. For triangulated planar graphs, this is easily reduced to linear: cutting pairs correspond to interior edges where both ends are exterior, and the necklace can be found, as in [1], with a left-first search that only advances neighbours of $S_{WY}\setminus B$. But for graphs that are not triangulated we need a few extra data structures. We sketch only some ideas for this here; details are in Section 5.

Globally, we keep track of the corners $X,U,W$, and $Y$. For each interior vertex $w$ and every side $S_{ab}$, we keep a list ${\mathcal{V}}(w,S_{ab})$ of faces that contain $w$ as well as a vertex on $S_{ab}$. In these lists, we can look up quickly whether an interior vertex is face-adjacent to a side. Also, each face knows for each side which vertices it has on it. Finally, for each pair of sides $S_{ab}$ and $S_{cd}$, we store a list ${\cal P}(S_{ab};S_{cd})$ of faces that are incident to a vertex on $S_{ab}$ and a (different) vertex on $S_{cd}$, i.e., faces that connect cutting pairs.

This allows to test for Case 2.1.2 and Case 2.1.3 easily (“is ${\cal P}(S_{XU},S_{WY})$ resp. ${\cal P}(S_{WY},S_{YX})$ non-empty?”), and Case 2.1.1 and Case 2.1.5 are easily determined from the planar embedding. We keep ${\cal P}(S_{WY},S_{YX})$ in an order such that its first entry is the appropriate cutting pair in Case 2.1.3. To find a necklace, we scan the faces incident to $x_{1},\dots,x_{s}$. More precisely, we consider (for vertex $x_{i}$, presuming we know face $f_{i}$ already) each face $f$ in ccw order after $f_{i}$, and along face $f$ each vertex $w$ in ccw order after $x_{i}$, until we find vertex $B$ (then we are done) or a vertex that is face-adjacent to a vertex in $S_{WY}\setminus B$ (then this is $x_{i+1}$ and $f_{i+1}=f$ and we repeat). The running time for this is proportional to the degrees of vertices and faces that were scanned. We also need to update the data structures when recursing into a subgraph; here, we scan along all vertices (and their incident faces) that were in some necklace along which we cut the graph, or that became newly exterior.

Two crucial insights are needed to bound the running time. First, we need to scan vertices and faces only if they become incident to a side that they were not previously incident to. Second, once a vertex or face is incident to a side, it remains incident to it forever (for some suitable definition of “side”). The two combined mean that every vertex and face is scanned only a constant number of times, because there are only four sides to have incidences with, and the linear running time follows. In fact, we only scan vertices and faces that are incident to the outer face in some subgraph, which means that they will be incident to the path $P$ that we compute, and we have the following:

Theorem 2**.**

The Tutte path $P$ for Theorem 1, Lemma 1 or Lemma 2 can be found in linear time. More specifically, the running time is $O(\sum_{f\in F(P)}\deg(f))$.

3 Applications

A number of interesting properties of planar 3-connected graphs can be derived easily from the existence of ${T_{\mathrm{SDR}}}$-paths. In particular, every planar 3-connected graph has a spanning tree of maximum degree 3 [2] (a concept known in the literature as a 3-tree, but we prefer to use the term binary spanning tree to avoid confusion with maximal graphs of treewidth 3). Secondly, every planar 3-connected graph has a 2-walk, i.e., a walk that visits every vertex at least once and at most twice [9]. In Section 6 and Section 7, we show that, using Lemma 1, these can be found in linear time; this was known for binary spanning trees [21, 3], but for 2-walks the previous best running time was $O(n^{3})$ [20].

Theorem 3**.**

Let $G$ be a 3-connected plane graph with exterior vertex $X$. Then $G$ has a binary spanning tree $T$ that can be found in linear time. Moreover, when rooting $T$ at $X$, a vertex $v$ has two children only if it is an interior vertex and part of a cutting triplet $\{v,w,x\}$ of $G$; one of the subtrees of $v$ contains exactly the vertices interior to $\{v,w,x\}$.

Theorem 4**.**

Let $G$ be a 3-connected plane graph with exterior vertex $X$. Then $G$ has a 2-walk $P$ that can be found in linear time. Moreover, $P$ visits $X$ exactly once, and it visits a vertex $v$ twice only if $v$ is part of a separating triplet.

4 Additional proof details

Throughout the text, we have left some of the more obvious details to the reader. This section provides detailed proofs for some of these results.

4.1 Arguing corner-3-connectivity

We defined corner-3-connectivity via the corner stellation, but to show it for some subgraph $G^{\prime}$, we used one of the following two arguments: Either $G^{\prime}$ was defined as the graph inside some noose ${\mathcal{N}}$, and the outer face of $G^{\prime}$ was a simple cycle, or we already knew that $G^{\prime}$ was corner-3-connected for some corners, but we claimed that we can omit some of the corners since there are no cutting pairs near them. We now formally prove that this is correct.

Lemma 3**.**

Let $G$ be a graph that satisfies $c3c(X,U,W,Y)$ for some corners $X,U,W,Y$. Let $H$ be the graph inside a noose ${\mathcal{N}}=\langle x_{1},\ldots,x_{k}\rangle$ that visits the outer face at most once. Let $H^{+}$ be the graph obtained from $H$ by adding some edges along ${\mathcal{N}}$. If the outer face of $H^{+}$ is a simple cycle, then $H^{+}$ satisfies $c3c(\{x_{1},\dots,x_{k}\}\cup{\mathcal{C}})$, where ${\mathcal{C}}\subseteq\{X,U,W,Y\}$ are the corners inside $\mathcal{N}$.

Proof.

See Fig. 7 for an illustration of this proof. Modify the noose $\mathcal{N}$ such that it also goes through the vertices in ${\mathcal{C}}$ while going through the outer face. Now the noose $\mathcal{N}$ contains vertices $\langle x_{1},\dots,x_{\ell}\rangle$, where ${\mathcal{C}}=\{x_{k+1},\dots,x_{\ell}\}$.

Consider the corner stellation $H^{\mathrm{s}}$ of $H^{+}$ with respect to $\{x_{1},\dots,x_{\ell}\}$; with our modification, noose $\mathcal{N}$ is a noose in $H^{\mathrm{s}}$ as well. Observe that any pair $x_{i},x_{j}$ is connected by three vertex-disjoint paths in $H^{\mathrm{s}}$: two along the outer face of $H^{+}$ and one via the stellation vertex $s$. Likewise, $s$ has three vertex-disjoint paths to any $x_{i}$ in $H^{\mathrm{s}}$: Use edges $(s,x_{i-1}),(s,x_{i}),(s,x_{i+1})$, and the outer face cycle of $H^{+}$. So if $H^{\mathrm{s}}$ had a cutting pair $\{v,w\}$, then all of $s,x_{1},\dots,x_{\ell}$ are in one cut component of $\{v,w\}$. In particular, therefore $v\neq s\neq w$ since $s$ has no other neighbors. Let $C$ be some cut component of $\{v,w\}$ that does not contain $s$. Then $C-\{v,w\}$ contains no vertices in $s,x_{1},\dots,x_{\ell}$, so all of them are strictly inside $\mathcal{N}$ and have no neighbor (in $G$) outside $\mathcal{N}$. Therefore, $\{v,w\}$ is also a cutting pair of $G^{\mathrm{s}}$, a contradiction. ∎

Lemma 4**.**

Assume that $G$ satisfies $c3c(c_{1},\dots,c_{k})$ for some $k\geq 4$. Assume that no cutting pair $\{v,w\}$ of $G$ satisfies $v,w\in S_{c_{1}c_{3}}$. Then $G$ also satisfies $c3c(c_{1},c_{3},\dots,c_{k})$.

Proof.

Let $G^{\mathrm{s}}$ be the corner stellation with respect to all corners $c_{1},\dots,c_{k}$. To prove the claim, we must show that $G^{\mathrm{s}}\setminus(c_{2},s)$ (where $s$ is the stellation vertex) is also 3-connected. Assume that it is not, in case of which it has a cutting pair $\{v,w\}$ that is necessarily also a cutting pair of $G$. Let $C_{1},C_{2}$ be two cut components of $\{v,w\}$ in $G$ (there are only two since $G$ is internally 3-connected). Since $G^{\mathrm{s}}$ is 3-connected, both $C_{1}^{+}$ and $C_{2}^{+}$ contain a corner other than $v,w$. This implies that both $v$ and $w$ lie on the outer face of $G$. Since they are not both in $S_{c_{1}c_{3}}$, therefore both outer face paths between $v$ and $w$ (and hence both $C_{1}^{+}$ and $C_{2}^{+}$) contain at least one corner other than $c_{2},v,w$. Therefore, $C_{1}^{+}$ and $C_{2}^{+}$ are connected in $G^{\mathrm{s}}\setminus(c_{2},s)$ by going from those corners to $s$. So $\{v,w\}$ is not a cutting pair of $G^{\mathrm{s}}\setminus(c_{2},s)$, a contradiction. ∎

4.2 Combining $T_{\mathrm{int}}$-paths

We generally obtained a $T_{\mathrm{int}}$-path by combining two or more $T_{\mathrm{int}}$-paths of subgraphs, possibly omitting some edges from these paths (but in such a way that the combination is a simple path). We now formally verify that this indeed gives a $T_{\mathrm{int}}$-path.

Lemma 5**.**

Let $\mathcal{N}$ be a noose of $G$ and let $H_{i}$ and $H_{o}$ be the graphs inside and outside $\mathcal{N}$, respectively. Let $P_{i}$ and $P_{o}$ be $T_{\mathrm{int}}$-paths of $H_{i}$ and $H_{o}$, respectively. Let $P\subseteq P_{i}\cup P_{o}$ be a simple path that visits all vertices of $P_{i}\cup P_{0}$. Then $P$ is a $T_{\mathrm{int}}$-path of $G$.

Proof.

Since $P_{i}$ and $P_{o}$ visit all their respective exterior vertices, $P$ visits all exterior vertices and all vertices in $\mathcal{N}$.

Consider a $P$-bridge $C$. It cannot have vertices both strictly inside and strictly outside $\mathcal{N}$, since all vertices on $\mathcal{N}$ are visited by $P$ and $C$ is connected. So $C$ is a $P$-bridge of $H_{i}$ or $H_{o}$, and can inherit the representative it received from there. No representative is used twice since the representatives of $P_{i}$ and $P_{o}$ were strictly inside/outside $\mathcal{N}$, respectively. Also, no representative is on the outer face, since none of $P_{i}$ and $P_{o}$ were. ∎

4.3 Existence of necklaces

We must argue that a suitable necklace exists in Case 2.1.5.

Lemma 6**.**

If none of Case 2.1.1,2.1.2,2.1.3,2.1.4 applies, then $G$ has a simple interior $B$-necklace for $B\in\{U,W\}$.

Proof.

It is easy to see that a $B$-necklace exists; for example, we can take all neighbors of $S_{WY}\setminus\{B\}$ that are not on the right side, enumerating them in order from $Y$ towards $W$ and in ccw order at each vertex. We claim that the resulting necklace $Y_{X}{=}x_{0},\dots,x_{s}{=}B$ is interior. For if $x_{i}$ (for some $0<i<s$) were exterior, then $x_{i}$ and its neighbor $t_{i}\in S_{WY}$ would form a cutting pair separating $x_{0}$ and $x_{s}$. So if $x_{i}$ is on the left side, then we are in Case 2.1.2, and if it is on the top side, then we are in Case 2.1.3 (which can be applied since $(x_{i},t_{i})$ is an edge).

Consider the curve ${\mathcal{C}}$ defined by this necklace. If ${\mathcal{C}}$ visits a vertex twice, then shortcut the necklace by removing the part between the two visits. If ${\mathcal{C}}$ crosses itself, say $x_{i-1}{-}x_{i}$ intersects $x_{j-1}{-}x_{j}$ for $i<j$, then we can immediately go from $x_{i-1}$ to $x_{j}$ with a curve along ${\mathcal{C}}$, thereby removing $x_{i},\dots,x_{j-1}$ from the necklace. So the shortest possible necklace that uses only vertices in $\{x_{0},\dots,x_{s}\}$ is simple and interior. ∎

5 Linear-time complexity for 3-connected graphs

We now explain how to find the ${T_{\mathrm{int}}}$-path in linear time for 3-connected graphs, i.e., prove Theorem 2. We first argue how to bound the time spend on recursions, once we know which case applies and have found the subgraphs. (This is the easier part.) Next, we explain how to store cutting pairs (for determination of cases) and how to store adjacencies to sides (for determination of the necklace); these data structures are not complicated, but arguing that their updates take overall linear time is lengthy.

5.1 Preliminaries.

A few notations will be useful. First, set $D_{F}:=\sum_{f\in F(P)}\deg(f)$; we aim to show that the running time is $O(D_{F})$. Next, set $D_{V}:=\sum_{v\in V(P)}\deg(v)$, where $V(P)$ are the vertices of $P$. Since every vertex has an incident interior face, we have $D_{V}\leq D_{F}$; it hence suffices to argue a running time of $O(D_{V}+D_{F})$. Next, set ${\mathcal{G}}$ to be the set of all subgraphs that we used in some recursion. Let $G^{+}$ be the graph that we would get if we inserted into $G$ all the edges that were used as virtual edge in some recursion. Note that $G^{+}$ is still a planar graph.

Let $V_{X}$ be the set of vertices of $G$ that were exterior in some subgraph $G^{\prime}\in{\mathcal{G}}$. Note that, when recursing into $G^{\prime}$, we obtained a Tutte path of $G^{\prime}$ that visits all exterior vertices in $G^{\prime}$. When combining Tutte paths of subgraphs, the resulting path always visits the same set of vertices. So $V_{X}\subseteq V(P)$, and in particular $|V_{X}|\leq|V(P)|\leq D_{V}$.

Let $E_{X}$ be the set of edges of $G^{+}$ (i.e., possibly including some virtual edges) that were exterior in some subgraph $G^{\prime}\in{\mathcal{G}}$. The ends of such edges necessarily belong to $V_{X}$. Since $G^{+}$ is planar, we have $|E_{X}|\leq 3|V_{X}|-6\in O(D_{V})$.

For any subgraph $G^{\prime}\in{\mathcal{G}}$, let $E_{X}(G^{\prime})$ be the edges of $G^{+}$ that are exterior in $G^{\prime}$, but were not exterior (or did not exist) in the parent graph that we recursed from. Note that $|E_{X}(G^{\prime})|\geq 1$ in all cases. The work done in the recursion for $G^{\prime}$ (not including the time to determine the case or to find the necklace) is $O(|E_{X}(G^{\prime})|)$, because we must update the planar graph embedding at the places where we split the parent graph to obtain $G^{\prime}$ (details are given below). Observe that $\sum_{G^{\prime}\in{\mathcal{G}}}|E_{X}(G^{\prime})|\leq 2|E_{X}|$, because any edge becomes exterior only once, and then belongs to at most two subgraphs that we recurse in. Therefore, the time to handle each recursion (excluding the time to find the case and the necklace) is $O(D_{V})$. The bottleneck for proving Theorem 2 is hence to show how we can test for cutting pairs and find leftmost necklaces in overall time $O(D_{V}+D_{F})$.

5.2 Data structures.

We presume that with the planar embedding we obtain a standard data structure with the following:

• •

Every vertex $v$ knows whether it is exterior, and has a list $L(v)$ of its incident faces and edges in ccw order. This list is circular if $v$ is interior, and begins and ends with the exterior edges at $v$ otherwise. (There are exactly two such edges since all graphs in ${\mathcal{G}}$ are 2-connected.)

• •

Every interior face $f$ has a list $L(F)$ of its incident vertices and edges in ccw order.

One could demand a similar list for the outer face, but since our outer face changes frequently we will not do this. Observe that even without such a list we can walk along the outer face in ccw order, presuming we know at least one vertex $v$ on it, by traversing the last edge in $L(v)$ and recursing from its other end.

• •

Vertices, edges and faces are cross-linked, i.e., each vertex/edge/face knows all its occurrences in the lists $L(v)$ and/or $L(F)$.

Setting up this data structure is standard material and will not be explained here; see, for example, [5, 16]. To find cutting pairs and leftmost necklaces efficiently, we need to store more information as follows; see Fig. 8.

• $\mathcal{C}$:

We store a circular list ${\mathcal{C}}$ of four references $c_{0},c_{1},c_{2},c_{3}$ to vertices in $G$; these are the corners of $G$ enumerated in ccw order. We also know which of these corners is $X$. Note that the “start corner” $X$ may change when recursing in a subgraph (consider for example graph $G_{0}^{+}$ in Case 2.1.5), but we keep the vertex in the same place in ${\mathcal{C}}$ and only change the reference to $X$; see Fig. 8(a)

We do not explicitly store the sides; any side is uniquely determined by its corners. We use $S_{i}$ to denote the side from $c_{i}$ to $c_{i+1}$ (addition for corners is always modulo 4). It is important that sides are defined via entries in ${\mathcal{C}}$ (not via references to vertices) due to the following. When recursing in a subgraph $G^{\prime}$, we may replace an entry $c_{i}$ in ${\mathcal{C}}$ by a new vertex $c_{i}^{\prime}$; we call $c_{i}^{\prime}$ the corner corresponding to $c_{i}$ . With this, $S_{i}$ now automatically refers to the side that starts at $c_{i}^{\prime}$ in $G^{\prime}$ (we call this the corresponding side in $G^{\prime}$).

We call an interior face $f$ incident to side $S_{i}$ if it contains a vertex $v$ on $S_{i}$. We order the interior faces incident to side $S_{i}$ as follows: We start with the interior faces around $c_{i}$ in clockwise order, hence ending with the interior face incident to $c_{i}$ and its neighbor $u_{1}$ on $S_{i}$. Next come the remaining interior faces around $u_{1}$, in clockwise order, ending with the interior face incident to $u_{1}$ and its neighbor $u_{2}\neq c_{i}$ on $S_{i}$. Continue until we reach vertex $c_{i+1}$. With this definition in place, we need the following three lists at vertices, faces, and sides:

• $\mathcal{F}$:

For every interior face $f$ and every $i=0,\dots,3$, we store a (possibly empty) list ${\mathcal{F}}(f,i)$ of vertices that are on $f$ and on side $S_{i}$. There are at most two such vertices per side by corner-3-connectivity, else the two non-consecutive vertices of $f$ on $S_{i}$ would form a cutting pair within $S_{i}$; see Fig. 8(b).

This obvious fact is the crucial ingredient for our data structures, because with this many checks can be done in constant time that otherwise would take longer. For example, given a vertex $w$, we can check in constant time which side(s) (if any) it lies on. Namely, take an arbitrary interior face $f$ at $w$ and inspect ${\mathcal{F}}(f,i)$ for each $i=0,\dots,3$. Since each list has constant size, we can check in constant time whether $w$ is in it, hence whether it lies on $S_{i}$.

• $\mathcal{V}$:

For every interior vertex $w$ and every $i=0,\dots,3$, we store a (possibly empty) list ${\mathcal{V}}(w,i)$ of interior faces $f$ that contain $w$ as well as a vertex on side $S_{i}$. The list is sorted by the order in which these faces are incident to side $S_{i}$; see Fig. 8(c).

Note that ${\mathcal{V}}(w,i)$ allows us to check in constant time whether $w$ is face-adjacent to corner $c_{i}$, because any interior face that contains $w$ and $c_{i}$ would have to be the first entry in ${\mathcal{V}}(w,i)$ by the order of faces incident to $S_{i}$.

• $\cal P$:

For any two distinct sides $S_{i},S_{j}$ we store a list ${\cal P}(i,j)$ of interior faces that contain a cutting pair $\{v,w\}$ of $G$, with $v$ on side $S_{i}$ and $w$ on side $S_{j}$. This list is sorted by the order in which these faces are incident to side $S_{i}$; see Fig. 8(d).

Note that ${\cal P}(j,i)$ is the reverse of ${\cal P}(i,j)$, but it will be convenient to store both of them. We assume that any face $f$ knows of all its occurrences in some list ${\cal P}(i,j)$; this is only a constant overhead per face.

Initialization. We argue how to initialize the data structure in $O(n)$ time; this then also shows that it uses $O(n)$ space. We assume that we are given references to the corners, and hence initialize ${\mathcal{C}}$ by walking around the outer face. For all other data structures, we call scan_side($i,c_{i},c_{i+1}$) for $i=0,\dots,3$. See Algorithms 1 and 2. This visits all exterior vertices, scans all their incident interior faces, and updates all the lists as needed. One verifies that, since we scan along a side in ccw order, the lists automatically receive the correct order. We should also mention that the conditions in lines 1 and 1 (explained below) are always true for the initialization and can be tested in constant time for later recursions. Condition 2 can be tested in constant time per vertex by marking (before starting any scan for any side) all vertices that will be scanned; in the initialization, this is all vertices.

The operation of scanning a vertex $v$ takes $O(\deg(v))$ time plus the time to scan its incident interior faces which we will count with those faces. The operation of scanning a face $f$ takes $O(\deg(f))$ time. In total, the running time of the initialization is therefore proportional to the degrees of vertices and faces that are incident to the outer face, which is $\Theta(D_{V}+D_{F})$ since $P$ visits all exterior vertices.

5.3 Updating

We first give some general rules for how to update the data structure, and then fill in for each individual case some case-dependent details. Assume that we recurse into some subgraph $G^{\prime}$ of $G$. Where useful, we will use “primed” versions (such as ${\mathcal{V}}^{\prime}$ and $S^{\prime}_{i}$) for the data structures and properties of $G^{\prime}$.

Each case will state which corner $Z^{\prime}$ of $G^{\prime}$ corresponds to which corner $Z$ of $G$, i.e., takes the entry of $Z$ in ${\mathcal{C}}$; this defines corresponding sides. We will do this so that the following holds.

Property 1**.**

Assume that a vertex $v$ is incident to a side $S_{i}$ in $G$ and belongs to some subgraph $G^{\prime}$ that we recurse in. Then $v$ is incident to the corresponding side $S_{i}^{\prime}$ of $G^{\prime}$.

We cannot afford to fully copy the graph structure (i.e., lists $L(v)$ and $L(f)$) from $G$ into $G^{\prime}$ since this would be too slow. Instead, observe that $G^{\prime}$ is in all cases defined as the inside of some noose ${\mathcal{N}}$ of $G$. At every vertex $v$ traversed by noose ${\mathcal{N}}$, we split the incidence list $L(v)$ into the two parts at the faces traversed by ${\mathcal{N}}$; these faces become the outer face of $G^{\prime}$ and hence this sets the lists up correctly. For any interior face $f$ traversed by ${\mathcal{N}}$, we might insert a virtual edge along $f$; if we do so, then we assume that the part inside ${\mathcal{N}}$ (hence belonging to $G^{\prime}$) inherits the reference $f$ and the (suitably shortened) list $L(F)$. (We assume that here, as in many of the other operations below, we store where the lists were cut, and keep a reference to the rest, so that we can restore the rest when returning from the recursion in $G^{\prime}$ and recursing into a different subgraph of $G$. We view this record as “belonging” to the (possibly virtual) edge that newly became an exterior edge of $G^{\prime}$; the final set of such edges form a planar graph, thus this overhead takes space $O(n)$ overall.)

Subgraph $G^{\prime}$ inherits the lists ${\mathcal{V}}$, i.e., ${\mathcal{V}}^{\prime}(w,i):={\mathcal{V}}(w,i)$ for all vertices $w$ that are in $G^{\prime}$, and no time is spent on creating these since we simply keep the same lists. We need to argue that this does not create false positives for an interior vertex $w$ of $G^{\prime}$. Recall that ${\mathcal{V}}(w,i)$ stores interior faces $f$ that contain $w$ as well as a vertex on $S_{i}$. Clearly, $f$ still contains $w$ in $G^{\prime}$, and it does contain a vertex on the corresponding side $S_{i}^{\prime}$ of $G^{\prime}$ because of the following property of our cases:

Property 2**.**

Assume that an interior face $f$ is incident to side $S_{i}$ in $G$, and $f$ (or some part of $f$ obtained by dividing $f$ along a virtual edge) is an interior face of some subgraph $G^{\prime}$ that we recurse in. Then $f$ is incident to the corresponding side $S_{i}^{\prime}$ of $G^{\prime}$.

This property is quite obvious for a face $f$ that was not divided, because then all vertices of $f$ must also belong to $G^{\prime}$, and by Property 1 such a vertex remains on all sides that it was on in $G$. This property is not at all obvious for faces that were divided by insertion of virtual edges; we will argue this for each case below.

Subgraph $G^{\prime}$ also inherits the four lists ${\mathcal{F}}(f,i)$ for any interior face $f$. If $f$ was not divided by a virtual edge, then this does not add false positives since any vertex that $f$ had on side $S_{i}$ of $G$ also is in $G^{\prime}$ (else $f$ would have been divided) and on side $S_{i}^{\prime}$ (by Property 1). If $f$ was divided by a virtual edge $(v,w)$, then we must change ${\mathcal{F}}(f,i)$, where $S_{i}$ is the side that contains $(v,w)$. We know $i$ from each case below, and set ${\mathcal{F}}(f,i):=\{v,w\}$. (We should also keep a copy of the prior list ${\mathcal{F}}(f,i)$ until we are done with updating the entire data structure, because in some of the tests below we need to use the lists as they were in $G$. This is only constant space overhead.)

The initialization of the ${\cal P}$-lists for $G^{\prime}$ will depend on each case, but as will be seen, they are either empty or inherited from $G$ (with minor modifications). Furthermore, all cutting pairs of $G$ that also exist in $G^{\prime}$ get copied over.

Scanning (parts of) sides. We have initialized the data structures of $G^{\prime}$ so that it has no false positives (entries that should not be there), but it may be missing some entries since some vertices may be new to a side. To add these missing entries, we scan (parts of) each side $S^{\prime}_{i}$ in such a way that all vertices that are new to $S^{\prime}_{i}$ are guaranteed to be scanned, i.e., we call scan_side($i,c,d$) for some $i\in\{0,\dots,3\}$ and some vertices $c,d\in S^{\prime}_{i}$. (Each case listed below will explain exactly what needs to be scanned.) We should mention here that the pseudocode needs a few minor modifications to obtain the correct order for ${\mathcal{V}}$ and ${\cal P}$; we will explain this below. To make this feasible, we need that $c,d$ are not arbitrary, and that the following holds in our cases:

Property 3**.**

Assume that, when recursing from $G$ to a subgraph $G^{\prime}$, we scan along a part $S^{\prime}_{cd}$ of some side $S^{\prime}_{i}$. Then, one of the following holds:

1. (i)

We scan the entire side (i.e., $c=c^{\prime}_{i}$ and $d=c^{\prime}_{i+1}$), and the vertices between $c$ and $d$ (if any) were not on side $S_{i}$, or 2. (ii)

$d=c^{\prime}_{i+1}$, $(c,d)$ is an edge, $c$ was on $S_{i}$ in $G$, and $d$ was not on $S_{i}$ in $G$.

It is worth pointing out that the second situation is needed in only one case (for graph $G_{b}^{+}$ in Case 2.1.3).

Recall Algorithm 1 for scanning sides. We exempt in line 1 a vertex $v$ from scanning if it was already on side $S_{i}$ before. (By Property 3, this can happen only to $c$ or to $d$, because all other vertices were new to the side.) First, note that this can be tested in constant time by inspecting ${\mathcal{F}}(f,i)$ (using the lists as they were in $G$) for some interior face $f$ incident to $v$. If $v$ was already on side $S_{i}$ in $G$, then scanning it would not add $v$ or an interior face $f$ incident to $v$ to any ${\mathcal{V}}$-list or ${\mathcal{F}}$-list, because all these lists already contained it in $G$ and were inherited. Also, any cutting pair $\{v,w\}$ of $G^{\prime}$ where both $v$ and $w$ already existed on their sides in $G$ was in some ${\cal P}$-list of $G$ and was inherited by $G^{\prime}$. If $w$ did not exist on its side (say $S_{j}$) in $G$, then we will scan at $w$ from side $S_{j}^{\prime}$, find the cutting pair then, and update ${\cal P}(i,j)$ as well due to line 2. So there is no need to scan at $v$.

We likewise exempt in line 1 a face $f$ from scanning if it was already on the side, i.e., if the list ${\mathcal{F}}(f,i)$ in $G$ was non-empty. As above, one argues that this will mean no missed entries in the data structure. Avoiding these face scans is the crucial insight to bring the running time down to linear.

Lemma 7**.**

The total time to scan sides of subgraphs of $G$ is $O(D_{V}+D_{F})$.

Proof.

When scanning sides (for one subgraph $G^{\prime}$), we may spend time on vertices that end up not being scanned, due to line 1. By Property 3, there are at most two vertices on each side, hence $O(1)$ in total. We count this as overhead to the time $O(|E_{X}(G^{\prime})|)$ that we budgeted for $G^{\prime}$ earlier, and do not consider it further here.

Likewise, we spend some time on interior faces that were incident to a scanned vertex $v$, but already incident to the side of $v$ and therefore end up not being scanned due to line 1. We spend $O(1)$ time per such face, hence $O(\deg(v))$ time per scanned vertex $v$. We count this as overhead to the time that we budget for scanning $v$, and do not consider it further here.

So we must only bound the time spend on scanning faces and vertices that were actually new to the side. Every vertex is incident to at most 4 sides, and since it never loses a side incidence by Property 1 and gains one with every scan by the rule in line 1, it is scanned $O(1)$ times.

Consider a face $f$ that gets scanned at least once. During some recursion, face $f$ may get split into pieces, which in turn can get split into more pieces in later recursions; let $f_{1},\dots,f_{k}$ be the (disjoint) pieces of $f$ that do not get split further, and note that $k\leq\deg(f)-2$ since at the worst $f$ gets split into $\deg(f)-2$ triangles.

Consider some part $f^{\prime}$ of $f$ (possibly $f^{\prime}=f$) that gets scanned, which takes $O(\deg(f^{\prime}))$ time. Let $f_{i_{1}},\dots,f_{i_{\ell}}$ (with $i_{j}\in\{1,\dots,k\}$) be the pieces of $f$ that belong to $f^{\prime}$. Therefore $\sum_{j=1}^{\ell}\deg(f_{i_{j}})\geq\deg(f^{\prime})$. So it suffices to account for the work if we assign $O(\deg(f_{i_{j}}))$ work to each piece $f_{i_{j}}$. During the scan of $f^{\prime}$, it acquires a new side incidence due to the rule in line 1. Crucially, by Property 2, this side incidence also exists in each $f_{i_{j}}$. Therefore, we count $O(\deg(f_{i_{j}}))$ time only if $f_{i_{j}}$ acquires a new side incidence, which happens at most 4 times. In consequence, the total time spent on scanning all the pieces of $f$ is $O(\sum_{i=1}^{k}\deg(f_{i}))=O(\deg(f)+k)=O(\deg(f))$.

If a vertex or face gets scanned, then it either was incident to the outer face already (as is the case in the initialization step), or it became incident to one more side, hence incident to the outer face. So it is in $V_{X}$ (in case of a vertex) or incident to a vertex in $V_{X}$ (in case of a face). As before, $V_{X}\subseteq V(P)$, so only vertices on $P$ or faces incident to $P$ get scanned and the running time is as desired. ∎

Maintaining the correct order. During side_scan(i,c,d), we possibly add entries to ${\mathcal{F}}(f,i)$, ${\mathcal{V}}(w,i)$, and ${\cal P}(i,j)$, for some interior face $f$, vertex $w$, and side $S_{j}$. The ${\mathcal{F}}$-lists have constant size and no particular order, but the other two lists must keep entries in the order in which faces are incident to side $S_{i}$. Property 3 is crucial for showing this: we usually scan nearly the whole side.

Assume first that we scan an edge $(c,c^{\prime}_{i+1})$, where $c^{\prime}_{i+1}$ is new to side $S^{\prime}_{i}$. Therefore, the existing entries in ${\mathcal{V}}^{\prime}(w,i)$ or ${\cal P}^{\prime}(i,j)$ are faces whose vertex on $S^{\prime}_{i}$ comes before $c^{\prime}_{i+1}$. So new entries (which must all be due to entry $c^{\prime}_{i+1}$ since all other vertices on $S_{i}^{\prime}$ are also on $S_{i}$) can simply be appended and we maintain the order.

Now, consider the case where we scan an entire side $S_{i}$, except that $c^{\prime}_{i}$ and/or $c^{\prime}_{i+1}$ may already have been adjacent to the side and then do not get scanned. If $c^{\prime}_{i+1}$ was not yet on $S_{i}$, then all new entries due to it should be appended to ${\cal P}(i,j)$. If $c^{\prime}_{i}$ was not yet on $S_{i}$, then all new entries due to it should be prepended (in reverse order). So we must only handle the case where both $c^{\prime}_{i}$ and $c^{\prime}_{i+1}$ were already on side $S_{i}$. If $S_{i}$ only consists of these two vertices, then no scanning happened and we are done; we may therefore assume that $S_{i}$ has at least three vertices. If list ${\cal P}(i,j)$ previously was empty, then we can simply scan. So by using the correct of the above approaches, we can create the correct order in ${\cal P}(i,j)$ as long as we ensure the following.

Property 4**.**

Assume that, when recursing from $G$ to a subgraph $G^{\prime}$, we scan an entire side $S^{\prime}_{i}$. Then either

1. (i)

$S^{\prime}_{i}$* contains at most two vertices, or* 2. (ii)

$c^{\prime}_{i}$* is new to $S_{i}$, or* 3. (iii)

$c^{\prime}_{i+1}$* is new to $S_{i}$, or* 4. (iv)

all lists ${\cal P}(i,j)$ are empty.

5.4 Details for the individual cases

We now finally go through all cases and fill in some case-dependent details. In particular, we must explain

1. (A)

how to test whether the case applies, and 2. (B)

how to obtain the information to create the subgraphs, especially the necklace in Case 2.1.5.

We must also explain for each subgraph $G^{\prime}$

1. (C)

which corners of $G^{\prime}$ correspond to which corners of $G$ so that Property 1 holds, 2. (D)

why no divided face loses a side incidence (Property 2), 3. (E)

how to initialize the ${\cal P}$-lists with the cutting pairs of $G$ that also exist in $G^{\prime}$, 4. (F)

which vertices may be new to sides, and that we can scan them by scanning sides in such a way that Property 3 and Property 4 holds.

In all cases, we continue with the notations and assumptions that were used in Section 2.1.

Case 2.1.1 and the substitution trick (Fig. 9).

1. (A)

It is straightforward to test whether Case 2.1.1 applies, since we know the outer face. 2. (B)

Observe that we only recurse into a subgraph if we apply the substitution trick. The subgraph is then $G^{\prime}:=G\setminus(W,Y)$, i.e., we remove the edge that was the right side. 3. (C)

The four corners of $G$ remain the same, so clearly Property 1 holds. 4. (D)

There are no divided faces. 5. (E)

Graph $G$ had no cutting pairs, so all ${\cal P}$-lists as empty. Hence Property 4 holds. 6. (F)

The only vertices that could be new to a side are the ones that shared a face with $(W,Y)$ in $G$. We should therefore scan along the entire side $S^{\prime}_{\operatorname{right}}$ of $G^{\prime}$.777“$\operatorname{right}$” is a shortcut for “the index $i$ such that $W=c_{i}$, i.e., side $S^{\prime}_{i}$ is the right side”. Since this can be looked up in $O(1)$ time from the location of $X$ and $Y$ in ${\mathcal{C}}$, we use this convenient shorthand for this and the other sides. All vertices in $S^{\prime}_{WY}\setminus\{W,Y\}$ are different from $W,Y$, hence new to $S^{\prime}_{\operatorname{right}}$.

Case 2.1.2 (Fig. 10).

1. (A)

If Case 2.1.1 does not apply, then we check the list of cutting pairs ${\cal P}(\operatorname{left},\operatorname{right})$; Case 2.1.2 applies if and only if it is non-empty. 2. (B)

Assuming ${\cal P}(\operatorname{left},\operatorname{right})$ is non-empty, let $f^{*}$ be its last entry (closest to $U$). Inspect ${\mathcal{F}}(f^{*},\operatorname{left})$ and ${\mathcal{F}}(f^{*},\operatorname{right})$ to find the vertices $v,w$; taking the ones closer to the bottom if there are two possibilities. Split $G$ into $G_{b}$ and $G_{t}$ along the noose through $v$, $f^{*}$, and $w$. If $(v,w)$ was not an edge of $G$, then also add the virtual edge $(v,w)$ to split $f^{*}$ into two parts, $f^{*}_{b}$ and $f^{*}_{t}$, in the subgraphs. 3. (C)

$G_{b}$ retains corners $U,W$, and $G_{t}$ retains corners $X,Y$; both gain new corners $v,w$. We replace corners in ${\mathcal{C}}$ (in this and all other cases) so that the retained corners keep their spots and the new corners fill the emptied spots in ccw order; then, Property 1 holds. 4. (D)

We must argue that $f^{*}$ loses no side incidences. Face $f^{*}$ was incident to both the left and the right side in $G$, and the same holds for both parts in both subgraphs. Further, $f^{*}$ is incident to the top side of $G$ if and only if $f^{*}_{t}$ is incident to the top side of $G^{+}_{t}$, and $f^{*}_{b}$ is always incident to the top side of $G^{+}_{b}$. Face $f^{*}$ is incident to the bottom side of $G$ if and only if $f^{*}_{b}$ is incident to the bottom side of $G^{+}_{b}$, and $f^{*}_{t}$ is always incident to the bottom side of $G^{+}_{t}$. So all side incidences are retained for the parts of $f^{*}$ (and possibly there are new ones). 5. (E)

We chose the cutting pair such that $G_{b}^{+}$ has no other cutting pairs, so the ${\cal P}$-lists for $G_{b}^{+}$ are initialized empty. Subgraph $G_{t}^{+}$ inherits the ${\cal P}$-lists from $G$, except that we remove face $f^{*}$ from ${\cal P}^{\prime}(\operatorname{left},\operatorname{right})$ if it had only one vertex each in ${\mathcal{F}}(f^{*},\operatorname{left})$ and ${\mathcal{F}}(f^{*},\operatorname{right})$. 6. (F)

Vertices $v,w$ may or may not be new to the bottom and the top side in $G_{t}$ and $G_{b}$, respectively. So we scan along the entire sides $S^{\prime}_{\operatorname{bottom}}$ in $G_{t}$ and $S^{\prime}_{\operatorname{top}}$ in $G_{b}$, which both consist of exactly the edge $(v,w)$. Since we scan along entire single-edge sides, Properties 3 and 4(i) hold.

Case 2.1.3 (Fig. 11).

1. (A)

To test whether Case 2.1.3 applies, we check whether ${\cal P}(\operatorname{right},\operatorname{top})$ is non-empty. Say it is, and let $f^{*}$ be its first face. By the order of faces in ${\cal P}(\operatorname{right},\operatorname{top})$, Case 2.1.3 can be applied if and only if $f^{*}$ does not contain $Y$, which we can learn in constant time from ${\mathcal{F}}(f^{*},\operatorname{top})$. 2. (B)

We find the cutting pair $\{y,w\}$ from ${\mathcal{F}}(f^{*},\operatorname{top})$ and ${\mathcal{F}}(f^{*},\operatorname{right})$. If either of these two lists contains two vertices, then we use the one that is clockwise later on the outer face. The resulting pair $\{w,y\}$ satisfies the assumptions on the pair by the order of ${\cal P}(\operatorname{right},\operatorname{top})$. Split $G$ into $G_{b}$ and $G_{t}$ along the noose through $y$, $f^{*}$, and $w$. If $(y,w)$ was not an edge of $G$, then also split $f^{*}$ into two parts $f^{*}_{b}$ and $f^{*}_{t}$ in the subgraphs $G_{b}^{+}$ and $G_{t}^{+}$, respectively. We recurse in $G_{b}^{+}$ first, which tells us whether Case 2.1.3 or 2.1.3 applies, i.e., whether to insert $(y,w)$ into $G_{t}$ or not. In Case 2.1.3, we must further delete $w$ if it had degree 1 in $G_{t}$. 3. (C)

$G_{b}^{+}$ retains corners $X,U,W$ and adds $y$ as its fourth corner. In Case 2.1.3, $G_{t}^{+}$ retains corner $Y$, and adds $y$, $w$, and again $w$. In Case 2.1.3, $G_{t}$ retains corner $Y$ and adds $y$, its neighbor $z$ on the outer face, and the vertex $w$. In Case 2.1.3, $G_{t}^{\prime}$ retains corner $Y$ and adds $y$, $z$, and the neighbor $x$ of $w$ as corners. 4. (D)

Let us first consider $f^{*}_{b}$. Subgraph $G_{b}^{+}$ has corners $X,U,W$, and $y$, with the same left and bottom side as $G$. Face $f^{*}$ is adjacent to the top and right side of $G$, and the same holds for $f^{*}_{b}$ in $G_{b}^{+}$ due to vertex $y$. Face $f^{*}$ cannot have been adjacent to the left side of $G$ (else we would apply Case 2.1.2). If face $f^{*}$ was adjacent to the bottom side, then this must hence have happened at $W$, which makes $\{y,W\}$ a cutting pair. By our sorting of ${\cal P}(\operatorname{right},\operatorname{top})$ hence $w=W$, which means that $f^{*}_{b}$ then is also adjacent to the bottom side at $w$. So $f^{*}_{b}$ retains all side incidences that $f^{*}$ had.

As for $f^{*}_{t}$, this is involved in a recursion only if Case 2.1.3 applies (else it becomes part of the outer face). Here, the corners of $G_{t}^{*}$ become $\{Y,y,w,w\}$, which means that face $f_{t}^{*}$ touches all four sides since it is incident to $y$ and $w$. 5. (E)

Observe first the status of the ${\cal P}$-lists in $G$. ${\cal P}(\operatorname{left},\operatorname{right})$ was empty, else Case 2.1.2 would have applied. Any face in ${\cal P}(\operatorname{right},\operatorname{top})$ is either $f^{*}$ or belongs to $G_{t}$, by our choice of $f^{*}$. Any face in ${\cal P}(\operatorname{left},\operatorname{top})$ belongs to $G_{b}^{+}$ because $f^{*}$ has no vertex on the left side (else Case 2.1.2 would apply). Any face in ${\cal P}(\operatorname{bottom},\operatorname{top})$ is incident to either $U$ or $W$. Those incident to $U$ belong to $G_{b}$. If any are incident to $W$, then $w=W$, these faces belong to $G_{t}$, and the first of them is $f^{*}$. So if $w=W$, then we split ${\cal P}(\operatorname{bottom},\operatorname{top})$ at the occurrence of $f^{*}$. The ${\cal P}$-lists for $G^{\prime}$ can now be initialized according to these insights, inheriting the list into the appropriate subgraph (after removing $f^{*}$ as needed) and initializing all other ${\cal P}$-lists as empty. 6. (F)

In $G_{b}^{+}$, no new vertices become exterior, but $y$ is new to the right side. So we scan along edge $(w,y)$, which is part of the right side of $G_{b}^{+}$. Note that this fits Property 3(ii) and Property 4(iii).

For $G_{t}$, observe that $S^{\prime}_{wY}$ and $S^{\prime}_{Y{y}}$ are part of their corresponding sides and need not be scanned, but we scan the other two sides completely. In Case 2.1.3, this means that we scan $S^{\prime}_{yw}$ and $S^{\prime}_{ww}$; since $(y,w)$ is an edge, Properties 3(i) and 4(i) hold.

In Case 2.1.3, we scan along side $S^{\prime}_{yz}$ and $S^{\prime}_{zw}$. The former is a single edge, so Properties 3(i) and 4(i) hold. The latter may have three or more vertices, but all vertices other than $w$ were interior in $G$, hence are new to the side. So Property 3(i) and Property 4(ii) hold, since $z$ is new to $S^{\prime}_{zw}$. The argument is the same for Case 2.1.3, replacing ‘$w$’ by ‘$x$’.

Case 2.1.4. If Case 2.1.3 does not apply, then we test for Case 2.1.4 by taking the last element $f^{*}$ of ${\cal P}(\operatorname{left},\operatorname{top})$ and checking whether it contains $X$. The treatment of this case is symmetric to Case 2.1.3.

Case 2.1.5 (Figs. 12 and 13).

1. (A)

If none of the previous cases applies, then we test condition 2 by walking from $W$ along the outer face to find out whether to apply Case 2.1.5 or Case 2.1.5. 2. (B)

Assume first that we have to apply Case 2.1.5. We then have to find a leftmost necklace ${\mathcal{N}}=\langle x_{0}{=}X,\ldots,x_{s}{=}W\rangle$, which is non-trivial. We determine the vertices in order $x_{s},x_{s-1},\dots,x_{1}$. Now do a left-first-search, starting from $W$ (see, e.g., de Fraysseix et al. [6] for more on left-first searches). However, we search not only along edges, but also along all face-adjacent vertices, and we advance the search only from vertices that are face-adjacent to $Y$. See Algorithm 3. We do some special handling once we found a face containing $X$, because this face (which becomes face $f_{1}$ of the necklace) is already incident to the top side and (as we will see) should not get scanned.

Since we perform a left-first-search, we find the leftmost path (where ‘path’ allows for face-adjacencies) that connects $W$ to $X$ along vertices that are face-adjacent to $Y$. Any $W$-necklace also defines such a path, so we find the leftmost $W$-necklace.

We claim that the total time spent on finding ${\mathcal{N}}$ is no more than the time spent for updating the data structures for graph $G_{0}^{+}$ later. First, we spend $O(1)$ time per vertex that we found, hence $O(s)$ time overall. Since $G_{0}^{+}$ has $s$ new edges on the outer face, this is accounted for by the $O(|E_{X}(G_{0}^{+})|)$ time that we already counted for $G_{0}^{+}$. Second, we spend $O(\deg(f^{\prime}))$ time on scanning an interior face $f^{\prime}$ incident to vertex $x$. This happens only if $f^{\prime}$ contains neither $X$ nor $Y$, because otherwise we immediately find the next vertex of the necklace, and this was counted above. Since $x$ becomes a vertex on the top side of $G_{0}^{+}$, and $f^{\prime}$ contains neither $X$ nor $Y$ (and hence no vertex on the top side of $G$), it is newly adjacent to the top side and hence will get scanned during the update.

For Case 2.1.5, finding the $U$-necklace is done similarly, except that we search from the top downward, rather than from the bottom upward, which requires exchanging ‘clockwise’ by ‘ccw’ in the order of scan. See Algorithm 4. Recall that $x_{1}$ is not arbitrary in Case 2.1.5; it is the clockwise neighbor after $Y$ in $L(Y_{X})$, and we want the leftmost necklace containing it.

Since we are doing a (modified) right-first-search from $x_{1}$, we will find the leftmost necklace that includes $x_{1}$ as desired. The running time is $O(s)+O(\sum_{f^{\prime}}\deg(f^{\prime}))$, where we sum over all those interior faces $f^{\prime}$ that are incident to $x_{i}$ (for some $1\leq i\leq s-1$), and did not contain $U$ or $W$ (else we immediately have found the next vertex of the necklace). Any such face $f^{\prime}$ was hence not incident to the bottom side of $G$, but is incident to the side $\langle x_{s},\dots,x_{1}\rangle$ that becomes the corresponding side in $G_{0}^{+}$. So, as above, the time for the necklace scan is a constant overhead for the time for the scanning done to update the data structures.

Note that, in both cases, once we found the necklace $\langle x_{0},f_{1},x_{1},f_{2},\dots,f_{s},x_{s}\rangle$, we can also easily find the vertices $t_{i}$ for $i=1,\dots,s-1$ ($t_{0}$ and $t_{s}$ are determined from the case). Namely, to find $t_{i}$, let $f^{\prime}$ be the last face in ${\mathcal{V}}(x_{i},\operatorname{right})$ (this exists by choice of $x_{i}$); then, $t_{i}$ is one of the vertices in ${\mathcal{F}}(f^{\prime},\operatorname{right})$, using the one that comes ccw later on the outer face if there are two. This gives all the required information to determine the subgraphs to recurse in. 3. (C)

In Case 2.1.5, graph $G_{0}^{+}$ retains corners $X,U,W$, and adds $W$ again as a corner. In Case 2.1.5, graph $G_{0}^{+}$ retains corners $X,U$ (though their order “rotates”; $X$ was the top-left corner in $G$ but becomes the top-right corner in $G_{0}^{+}$) and adds $x_{1}$ and $x_{0}=Y_{X}$ as corners.

Any subgraph $G_{i}$ (for $i=1,\dots,s$ and in both Case 2.1.5 and 2.1.5) has four corners $x_{i-1},x_{i},t_{i},t_{i-1}$. We let $t_{i-1}$ take the place of $Y$ in ${\mathcal{C}}$ (with this, side $S^{\prime}_{t_{i},t_{i-1}}$ of $G_{i}$ corresponds to side $S_{WY}$ of $G$ as required for Property 1).

As always, corners that are neither retained nor explicitly assigned a place in ${\mathcal{C}}$ are filled in as to maintain the ccw order in ${\mathcal{C}}$. 4. (D)

We need to argue that any face $f$ that was divided in Case 2.1.5 does not lose any side incidences.

Assume first that some part of $f$ belongs to graph $G_{0}^{+}$ in Case 2.1.5. Any incidence of $f$ with the left or bottom side of $G$ is carried over since $G_{0}^{+}$ inherits these sides. So we only have to worry about incidences of $f$ with the right or top side of $G$. But $f$ cannot be incident to $Y$, else by Claim 1 it would have no part in $G_{0}^{+}$. So if $f$ was incident to the top side, then by 2 it must contain $X$, and it retains this vertex on the top side of $G_{0}^{+}$. Similarly, if $f$ was adjacent to the right side, then by 2 it must contain $W$, and it retains this vertex on the right side of $G_{0}^{+}$.

Assume next that $f$ has parts in graph $G_{0}^{+}$ in Case 2.1.5. Any incidence of $f$ with the left side of $G$ is carried over since $G_{0}^{+}$ inherits this side. Face $f$ cannot contain any vertex on the right side of $G$ (and in particular not $Y$ or $W$), else by Claim 1 it would have no part in $G_{0}^{+}$. So $f$ was not incident to the right side of $G$. Furthermore, if $f$ was incident to the top side of $G$, then this incidence happened at a vertex $z\neq Y$, so $z$ is also in $G_{0}^{+}$ and hence $f$ retains this side incidence. If $f$ was incident to the bottom side $(U,W)$ of $G$, then it is incident to $U$ which belongs to $G_{0}^{+}$, and so $f$ retains this side incidence.

Assume next that $f$ has parts in graph $G_{i}$ for some $1\leq i\leq s$, and $t_{i-1}=t_{i}$. We recurse in $G_{i}$ only if we apply the substitution trick, in case of which edge $(x_{i-1},x_{i})$ is removed. In particular, the face $f_{i}$ that was split along $(x_{i-1},x_{i})$ has no part in the graph $G_{i}^{+}:=G_{i}\cup(x_{i-1},t_{i})\cup(x_{i},t_{i})$ in which we recurse. So $f\neq f_{i}$, hence $f$ must have been split by $(x_{i-1},t_{i})$ or $(x_{i},t_{i})$ and $f$ contains $t_{i}$. Because $t_{i}$ becomes both the top-right and the bottom-right corner in $G_{i}^{+}$, face $f$ (which is also adjacent to one of the left corners $x_{i-1}$ and $x_{i}$) is incident to all four sides of $G_{i}^{+}$. In particular, no side incidence has been lost.

Assume finally that $f$ has parts in graph $G_{i}$ for some $1\leq i\leq s$, and $t_{i-1}\neq t_{i}$. (This in particular implies that we are in Case 2.1.5.) We argue that the side incidences are preserved for the resulting graph $G_{i}^{\sqsubset}$; then, they clearly also hold for $G_{i}^{\asymp}=G_{i}^{\sqsubset}\setminus(x_{i-1},x_{i})$ which has even larger sides.

Graph $G_{i}^{\sqsubset}$ has as its four sides the edges $(x_{i-1},t_{i-1})$, $(x_{i-1},x_{i})$, $(x_{i},t_{i})$, and path $S_{t_{i}t_{i-1}}$. Any face that was divided and has parts in $G_{i}^{\sqsubset}$ contains one of these three edges, hence immediately is adjacent to two corners and with them three sides, and we only need to worry about the one remaining side.

Assume that $f$ was divided by $(x_{i-1},x_{i})$, in case of which we only need to worry if $f$ had vertices on the right side of $G$. But then, by Claim 1, edge $(x_{i-1},x_{i})$ was not virtual and $f$ was not divided by this edge, a contradiction.

Assume that $f$ was divided by $(x_{i-1},t_{i-1})$, in case of which we only need to worry if $f$ had a vertex $z$ on the bottom side of $G$. If $z=U$, then $\{z,t_{i-1}\}$ would be a cutting pair by $t_{i-1}\neq t_{i}$, and we would have applied Case 2.1.2. So $z=W$, which implies $t_{i}=W$ (else we would have a cutting pair within the right side of $G$). So $f$ contains $t_{i}$, and hence is incident to all sides of $G_{i}^{\sqsubset}$.

Assume that $f$ was divided by $(x_{i},t_{i})$, in case of which we only need to worry if $f$ had a vertex $z$ on the top side of $G$. We claim that $f$ contains $Y$. Namely, if $z\neq Y$, then $\{z,t_{i}\}$ would be a cutting pair by $t_{i-1}\neq t_{i}$. Since we did not apply Case 2.1.3, therefore $f$ contains $Y$. This implies $t_{i-1}=Y$ (else we would have a cutting pair within the right side of $G$). So $f$ contains $t_{i-1}$, and hence is incident to all sides of $G_{i}^{\sqsubset}$. 5. (E)

Since we applied neither Case 2.1.2 nor Case 2.1.3, there are only two possible cutting pairs in $G$: The neighbors of $X$, if $\deg(X)=2$, and the neighbors of $Y$, if $\deg(Y)=2$.

If $\deg(X)=2$ or $\deg(Y)=2$, then the top side is not a single edge (else we would be in Case 2.1.1 or 2.1.2), so Case 2.1.5 applies. If $\deg(X)=2$, then its neighbors lie on the left and top side of $G$ (recall that we assumed $X\neq U$). So they belong to $G_{0}^{+}$ and form a cutting pair there. Initialize the appropriate ${\cal P}$-lists of $G_{0}^{+}$ with the interior face at $X$.

If $\deg(Y)=2$, then let $f_{Y}$ be its unique interior face. Let $Y_{X}$ be the neighbor of $Y$ on the top side; this exists because $X\neq Y$. If $Y=W$, then the neighbors of $Y$ would be $U$ and $Y_{X}$ and form a cutting pair on the left and top side. Since Case 2.1.4 does not apply, therefore $f_{Y}$ also contains $X$, and therefore $Y_{X}=X$ (else there is a cutting pair within the top side). But then $(X,U)$ must be an edge (else there would be a cutting pair within the left side), which means that the outer face is triangle and we are in the base case. So we know $Y\neq W$. Let $Y_{W}$ be the neighbor of $Y$ on the right side (possibly $Y_{W}=W$). Cutting pair $\{Y_{X},Y_{W}\}$ is inherited by the subgraph $G_{i}$ that contains the edge $(Y_{W},Y)$. We initialize the appropriate ${\cal P}$-lists of $G_{i}$ with the face $f_{Y}$.

All ${\cal P}$-lists not initialized by the above are initialized empty. 6. (F)

We scan in all subgraphs along all entire sides that contain newly exterior edges. Such sides are single-edge sides, with the exception of side $\langle x_{0},\dots,x_{s}\rangle$ (in Case 2.1.5) and $\langle x_{1},\dots,x_{s}\rangle$ (in Case 2.1.5). But $x_{1},\dots,x_{s-1}$ were interior and hence new to the side, so Property 3(i) holds. Also, the ${\cal P}$-lists are empty in Case 2.1.5, and in Case 2.1.5 all scanned sides are single edges or the side $\langle x_{1},\dots,x_{s}\rangle$, for which corner $x_{1}$ is new to the side. So Property 4 holds.

With this, we have explained how to do all steps in all cases. Note in particular that computing the necklace takes no more time than what was spent on scanning vertices and faces. Note further that Property 2 holds, hence the running time is linear (in the degrees of vertices and faces touched by the output path $P$) as desired.

6 Finding binary spanning trees

In this section, we prove Theorem 3, i.e., we show that any 3-connected planar graph has a binary spanning tree and it can be found in linear time. This was known before [3, 21], but we give a different proof here as a warm-up for Section 7 and because our binary spanning tree has some other interesting properties. We need a slightly stronger statement for the induction step to go through; in particular, we must be allowed to fix one leaf and one edge of the tree, as long as they are exterior to $G$.

Lemma 8**.**

Let $G$ be a plane graph with distinct vertices $X,Y$ on the outer face. Let $(U,W)\neq(X,Y)$ be an edge on the outer face. If $G$ is corner-3-connected with respect to $X,U,W,Y$, then it has a binary spanning tree $T$ such that when rooting $T$ at $X$

• •

$Y$* is a leaf and $(U,W)$ is an edge of $T$,*

• •

a vertex $v$ has two children in $T$ only if it is an interior vertex and part of a cutting triplet $\{v,w,x\}$ of $G$; moreover, one of the subtrees of $v$ contains exactly the vertices interior to $\{v,w,x\}$.

The tree $T$ can be found in linear time.

Proof.

We proceed by induction on the number of vertices, with an inner induction on the number of exterior vertices. In the base case, $G$ is a triangle, say $\langle X{=}U,W,Y\rangle$, and the path $\langle X,W,Y\rangle$ satisfies all properties. Now, assume that $n\geq 4$, and let $P$ be a ${T_{\mathrm{int}}}$-path from $X$ to $Y$ that contains $(U,W)$. By Theorem 2, we know that this path can be found in $O(\sum_{f\in F(P)}\deg(P))$ time.

We expand $P$ into a binary spanning tree $T$ by expanding each $P$-bridge $C$ into a subtree to be attached at its representative $\sigma(C)$; see Fig. 14. Formally, let $\{x,w,y\}$ be the attachment points of $C$, with $x=\sigma(C)$. Define $C^{+}$ as in the substitution trick, i.e., $C^{+}=G[C]\cup\{(x,w),(w,y)\}\setminus\{(x,y)\}$. This satisfies $c3c(x,w,y)$ and either has fewer vertices or has a bigger outer face. So, by induction, we can find a binary spanning tree $T_{C}$ in $C^{+}$ that, when rooted at $x$, has $y$ as a leaf and contains edge $(w,y)$; in particular, $w$ is the parent of $y$. Therefore , $T_{C}\setminus\{w,y\}$ is a spanning tree of $C\cup\{x\}$, and $x$ has only one child in it since it is exterior in $C^{+}$. We merge this subtree into $P$ at $x$. Repeating this for all $P$-bridges gives the final tree $T$. Clearly, the first property holds for $T$ since it held for $P$ already. If a vertex $v$ has two children in $T$, then necessarily $v=\sigma(C)$ for some $P$-bridge $C$. Therefore, $v$ is interior since $P$ is a ${T_{\mathrm{int}}}$-path. The rest of the second property follows immediately since one subtree of $v$ was obtained by merging $T_{C}\setminus\{w,y\}$.

As for the running time, observe that to find $T$ we first find a ${T_{\mathrm{int}}}$-path $P$ in $G$, then a ${T_{\mathrm{int}}}$-path $P_{C}$ for each $P$-bridge $C$, then a ${T_{\mathrm{int}}}$-path for each $P_{C}$-bridge, and so on recursively. Crucial for an efficient running time is that the time to find $P$ is proportional to $\sum_{f\in F(P)}\deg(f)$, i.e., faces that are strictly interior to $C$ do not contribute to this running time. Secondly, in the time for finding $P$ we can also initiate all the data structures as they are needed for $C^{+}$ (because this is the same graph as was used for the substitution trick). This saves having to re-initialize data structures by scanning when recursing into $C^{+}$. Therefore, exactly as for Theorem 2, one argues that all faces are scanned a constant time, giving an overall linear running time. ∎

7 Finding 2-walks

In this section, we prove Theorem 4, i.e., we show that every 3-connected planar graph has a 2-walk. This was originally shown by Gao and Richter [9] (with later improvements together with Yao [10, 11]). As in their proofs, we show instead the existence of a 2-circuit, i.e., a circuit in the graph that visits every vertex at least once and at most twice; a 2-walk can then be obtained by deleting one edge. Gao and Richter argue that a 2-circuit can be found by starting with a ${T_{\mathrm{SDR}}}$-path $P$ (with an extra edge added to close it to a circuit) and then recursively finding a 2-circuit in each $P$-bridge $C$ and attaching it at the representative of $C$.

While this is in spirit very similar to how we found binary spanning trees (Section 6), and so one would expect that this can be done in linear time using our algorithm for Tutte paths, there are some details that make this more complicated. In particular, Gao and Richter needed the ability to find a Tutte path in an internally 3-connected graph (which they called a circuit graph). Not every such graph satisfies corner-3-connectivity with at most 4 corners, so that we cannot apply Lemma 1 directly to find those Tutte paths. Therefore, we argue how to find the 2-circuits in linear time in two steps: first we argue how to find Tutte paths in internally 3-connected graphs in linear time, and then we repeat the proof of Gao, Richter, and Yao [10] to find 2-circuits, and argue that this also takes linear time.

7.1 Internally 3-connected graphs

We first expand Lemma 1 to graphs that are internally 3-connected. We consider here only the special case where $(X,Y)$ is an edge; the existence of a ${T_{\mathrm{SDR}}}$-path could be shown even in the absence of this edge, but if the edge exists, then we can restrict the representatives.

Lemma 9**.**

Let $G$ be a plane graph with distinct edges $(X,Y)$ and $(U,W)$ on the outer face. If $G$ is internally 3-connected, then it has a ${T_{\mathrm{SDR}}}$-path $P$ that begins at $X$, ends at $Y$, and contains $(U,W)$. Furthermore, $X$ and $W$ are not used as a representative. This path can be found in time $\sum_{f\in F(P)}\deg(f)$.

Proof.

We may assume that $X,U,W,Y$ occur in ccw order, the other case is symmetric. Consider first the left side (with respect to the four corners $X,U,W,Y$), and enumerate it as $\langle X{=}u_{0},u_{1},\dots,u_{\ell}{=}U\rangle$. There may now be cutting pairs within the left side. Let $i_{1}$ be minimal such that $\{u_{i_{1}},u_{j_{1}}\}$ is a cutting pair for some $j_{1}>i_{1}$, and assume that $j_{1}$ has been chosen maximal among all such cutting pairs. Repeating, for $k=2,3,\dots$, let $i_{k}\geq j_{k-1}$ be minimal such that $\{u_{i_{k}},u_{j_{k}}\}$ is a cutting pair for some $j_{k}>i_{k}$, and choose $j_{k}$ to be maximal among all such cutting pairs.

Now remove all these cutting pairs by removing, for each $k=1,2,\dots$ the cut component $C_{k}$ of $\{u_{i_{k}},v_{i_{k}}\}$ that does not contain the right side of $G$, and adding a virtual edge $(u_{i_{k}},u_{j_{k}})$ if it did not yet exist in $G$. Also set $\sigma(C_{k})=u_{j_{k}}$, i.e., assign as representative of $C_{j}$ the vertex of the cutting pair that is closer to $U$ along the left side. Note in particular that any cutting pair receives a distinct representative on the outer face, and $X$ will not be assigned as a representative. (Neither will $W$, since it is not on the left side.) See Fig. 15.

Similarly, remove all cutting pairs on the right side, replacing the cut component by a virtual edge and assigning as representative the vertex of the cutting pair that is closer to $Y$. Again, all representatives are distinct and $X,W$ will not be assigned as a representative.

Note that there can be no cutting pair within the top or bottom side, since these sides are single edges $(X,Y)$ and $(U,W)$. Thus, after having removed cutting pairs on the left and right side, the resulting graph $G^{\prime}$ is corner-3-connected with respect to $X,U,W,Y$. Apply Lemma 1 to $G^{\prime}$ to find a ${T_{\mathrm{int}}}$-path $P^{\prime}$ of $G^{\prime}$. None of the representatives used by $P^{\prime}$ conflicts with representatives assigned to cutting pairs, because the latter are all exterior.

If $P^{\prime}$ uses no virtual edge, then it can serve as the desired ${T_{\mathrm{SDR}}}$-path. Otherwise, for every virtual edge $(u_{i_{k}},u_{j_{k}})$ used by $P^{\prime}$, we replace this edge with a path through $C_{k}$. More precisely, recursively find a ${T_{\mathrm{SDR}}}$-path $P_{k}$ in $C_{k}^{+}:=G[C_{k}]\cup\{(u_{i_{k}},u_{j_{k}})\}$ that begins at $u_{i_{k}}$, ends at $u_{j_{k}}$, uses some other exterior edge, and does not use $u_{i_{k}}$ as a representative. This exists since $C_{k}^{+}$ is again internally 3-connected, and it does not use the (possibly virtual) edge $(u_{i_{k}},u_{j_{k}})$ since this edge connects the ends of $P_{k}$. Then substitute $P_{k}$ in place of edge $(u_{i_{k}},u_{j_{k}})$ in $P$.

It remains to analyze the time complexity. The bottleneck is finding the cutting pairs within each side. Let us assume that we have scanned the outer face of $G$ to mark each vertex with the side that it belongs to. Recall that we initiate the data structures for $G$ by scanning at each exterior vertex along all its incident faces. Say we do this scan in ccw order, and are currently scanning vertex $u_{i_{k}}$, starting clockwise after the outer face. If, during this scan, we encounter some vertex $u_{j_{k}}$ that is on the same side, then we have found a cutting pair that must be removed; we immediately split the graph at the currently scanned face (this removes the cut component $C_{k}$) and add the edge $(u_{i_{k}},u_{j_{k}})$ if it did not exist yet. Since we scanned faces in clockwise order, index $j_{k}$ automatically is maximized. Note that we had not yet scanned any faces that are interior to $C_{k}$, and we will not do so (at least not for the purpose of finding $P^{\prime}$) since $C_{k}$ is now removed from $G$.

Let $G^{\prime}$ be the resulting graph after removing all these cutting pairs. We can find the ${T_{\mathrm{int}}}$-path in time $O(\sum_{f\in F^{\prime}(P^{\prime})}\deg(f))$, where $F^{\prime}(P^{\prime})$ denotes the faces incident to $P^{\prime}$ in $G^{\prime}$. This proves the claim if we need not recurse into any cut component $C_{k}^{+}$. If we do recurse in $C_{k}^{+}$, then we must update the data structure for it. However, we are free to choose which exterior edge of $C_{k}^{+}$ must be visited by $P_{k}$; we can choose this to be the one incident to $u_{j_{k}}$ so that the entire outer face of $C_{k}^{+}$ (except for $u_{j_{k}}$) becomes the right side and can therefore inherit the side-marking that it had obtained in $G$. Then we recurse in $C_{k}^{+}$, which takes time at most $O(\sum_{f\in F_{k}(P_{k})}\deg(f))$, where $F_{k}(P_{k})$ are the faces in $C_{k}^{+}$ that are incident to $P_{k}$. This repeats for possibly many $P^{\prime}$-bridges, but since the interior faces of these $P^{\prime}$-bridges are distinct and $F^{\prime}(P^{\prime})\cup\bigcup_{k}F_{k}(P_{k})=F(P)$, the claim on the running time follows. ∎

7.2 2-circuits

We now use Lemma 9 to prove Theorem 4, i.e., to find 2-circuits in planar 3-connected graphs. The proof idea is exactly as in [10], we repeat it here to argue that it can be implemented in linear time. Again we need a slightly stronger statement to make the induction hypothesis work.

Lemma 10**.**

Let $G$ be an internally 3-connected plane graph with exterior vertices $X\neq W$. Then $G$ has a 2-circuit $P$ that visits $X$ and $W$ exactly once, and for which a vertex $v$ is visited twice only if $v$ is part of a separating triplet or a cutting pair. $P$ can be found in linear time.

Proof.

Let $Y$ and $U$ be the clockwise neighbor of $X$ and $W$ on the outer face. Find a ${T_{\mathrm{SDR}}}$-path from $X$ to $Y$ through $(U,W)$ using Lemma 9 and complete it to a circuit $P^{\prime}$ by adding edge $(X,Y)$. If $P^{\prime}$ visits all vertices of $G$, then we are done. Otherwise, there are some $P^{\prime}$-bridges, and we expand $P^{\prime}$ into a 2-circuit by recursively finding 2-circuits in the $P^{\prime}$-bridges and merging them into $P^{\prime}$. Formally, let $C$ be a $P^{\prime}$-bridge. We distinguish cases by whether $C$ has two or three attachment points.

$C$** has two attachment points $\{u_{i},u_{j}\}$; see Fig. 16(a):** Let us assume that $\sigma(C)=u_{j}$ is the representative of $C$. Let $C^{+}$ be the graph induced by $C\cup\{u_{i},u_{j}\}$ with edge $(u_{i},u_{j})$ added if it did not exist in $G$. We know that $C^{+}$ is internally 3-connected, therefore (as argued in [9]) $C^{+}\setminus u_{i}$ is a so-called plane chain of blocks (defined below).

$C$** has three attachment points $\{x,w,y\}$; see Fig. 16(b):** Let use assume that $\sigma(C)=x$ is the representative of $C$. Let $C^{+}$ be graph $G[C]\cup\{(x,w),(w,y),(y,x)\}$. We know that $C^{+}$ is 3-connected, therefore $C^{+}\setminus\{w,y\}$ is again a plane chain of blocks [9].

In both cases, hence the graph induced by $C\cup\sigma(C)$ is a plane chain of blocks. This means that it is the union of internally 3-connected graphs $B_{1},\dots,B_{k}$ where

1. (1)

$B_{1}$ contains $\sigma(C)$ (set $x_{1}:=\sigma(C)$), 2. (2)

for $\ell=2,\dots,k$ graphs $B_{\ell-1}$ and $B_{\ell}$ have one exterior vertex $x_{\ell}$ in common, 3. (3)

$\{x_{1},\dots,x_{k}\}$ are mutually distinct, and 4. (4)

the graphs are disjoint otherwise.

Recursively find 2-circuits $P_{1},\dots,P_{k}$ for $B_{1},\dots,B_{k}$ such that $P_{\ell}$ visits $x_{\ell}$ and $x_{\ell+1}$ exactly once (we set $x_{k+1}\neq x_{k}$ to be an arbitrary exterior vertex of $B_{k}$). Set $P_{C}$ to be $P_{1}\cup\ldots\cup P_{k}$, i.e., the union of the edges of the circuits. This has even degrees and hence is again a circuit. Vertex $x_{i}$, for $i>1$, is visited once each by $P_{i-1}$ and $P_{i}$, so visited twice by $P_{C}$. Observe that $x_{i}$ is a cut vertex of $G[C\cup\sigma(C)]$, hence either part of a cutting pair $\{x_{i},u_{i}\}$ or a cutting triplet $\{x_{i},w,y\}$. So $P_{C}$ is a 2-circuit of $G[C\cup\sigma(C)]$ that visits $\sigma(C)$ once and satisfies the requirements on vertices that are visited twice.

Adding $P_{C}$ to $P^{\prime}$ hence gives a circuit of $G$ that now visits all vertices in $C$ at least once. This new circuit visits vertex $\sigma(C)$ repeatedly, from $P^{\prime}$ and once from $P_{C}$. However, vertex $\sigma(C)$ was visited exactly once by $P^{\prime}$ initially, and it is used as a representative for only one $P^{\prime}$-bridge, so in the final circuit $P$ (obtained after applying the substitution for all $P^{\prime}$-bridges) $\sigma(C)$ is visited twice. Hence $P$ is a 2-circuit. Furthermore, since $\sigma(C)\neq X,W$ is an attachment vertex of $C$ and $P$-bridges have at most three attachment points, all properties hold.

It remains to analyze the running time, and in particular, how to find the chain of blocks efficiently. (We cannot afford to run the standard algorithms to find 2-connected components, because this takes linear time, which may result in overall quadratic time if we do it repeatedly in the recursions.) We explain this only for the case of three attachment points, the other case is similar. Assume as before that $C$ has attachment points $\{x,w,y\}$ with $\sigma(C)=x$. Recall that $x,w,y\in P$. We can therefore afford (as overhead to the time to find $P$) to scan those faces incident to $w,y$ that are inside $C^{+}$. So we can mark all interior vertices of $C^{+}$ that are face-incident to $w$, and (with a different mark) all interior vertices of $C^{+}$ that are face-incident to $y$. Any vertex that is marked from both $w$ and $y$ is a cut vertex of $C$ (or vertex $x$), and should hence become one of the vertices $x_{1},\dots,x_{k}$. These vertices are discovered in order if we scan the incident faces in order. All vertices that are face-incident to one of $w$ and $y$ (but not the other) become the outer face of $B_{1},\dots,B_{k}$. We have hence found the split into subgraphs to recurse in without scanning any faces inside those subgraphs. As for Lemma 9, one argues that all recursions overall take time proportional to the degree of all faces, which is linear. ∎

8 Outlook

In this paper, we improved on a very recent result that shows that Tutte paths in planar graphs can be found in quadratic time. We gave a different existence proof which leads to a linear-time algorithm. For 3-connected planar graphs, we obtain not only a Tutte path, but furthermore endow it with a system of distinct representatives, none of which is on the outer face. With this, we can also find 2-walks and binary spanning trees in 3-connected planar graphs in linear time.

The main remaining questions concern how to find Tutte path in other situations or with further restriction. For example, Thomassen [23] and later Sanders [17] improved Tutte’s result and showed that we need not restrict the ends of the Tutte path to lie on the outer face. These paths can be found in quadratic time [20]. But our proof does not seem to carry over to the result by Sanders, because the ends of the path crucially must coincide with corners of the graph. Can we find such a path in linear time?

Furthermore, the existence of Tutte paths has been studied for other types of surfaces (see, e.g., Kawarabayashi and Ozeki [15] and the references therein). Can these Tutte paths be found in polynomial time, and preferably, linear time?