Taking a Detour; or, Gioan’s Theorem, and Pseudolinear Drawings of Complete Graphs

Abstract

We describe a uniform approach to two known graph drawing results including Gioan’s theorem, stating that any two good drawings of a complete graph with the same rotation system are isomorphic up to Reidemeister moves of type 3, and a characterization of pseudolinear drawings of the complete graph via an excluded configuration: a bad $K_4$. Our approach yields a new and short self-contained proof of Gioan’s theorem, and a short proof of the pseudolinearity characterization using a previous result. As a bonus we obtain an extension of Gioan’s theorem to the family of graphs $K_n-M$, where M is a non-perfect matching in $K_n$, $n \ge 5$.

Appendices

A Crossings in Simultaneous Drawings

The usual perturbation/general position argument can be invoked to show that during the redrawings performed in this paper, different drawings of the same edge intersect only finitely often. To bound the number of slide moves in applications of the detour lemma asymptotically, we need a more precise statement, which the following theorem supplies.¹¹ The result applies to cone graphs, graphs that have a spanning vertex.

Theorem A.1

  Suppose $D_1$ and $D_2$ are good drawings of the same cone graph G on n vertices. Then there is a good drawing $D'_2$ isomorphic to $D_2$ such that

• (i) the vertices of G have the same locations in $D_1$ and $D_2'$,

• (ii) any two edges in $D_1 \cup D_2'$ are drawn the same way, or cross at most $O(n^6)$ times,

• (iii) if $D_1$ and $D_2$ have the same rotation system, then the ends of $e_1 \in D_1$ and $e_2 \in D_2$, both corresponding to the same edge $e\in E(G)$, are consecutive at their vertices in $D_1 \cup D_2'$, or e is drawn the same way in both $D_1$ and $D_2'$.

Proof

Since G is a cone graph, there is a vertex w such that S(w) spans all of G. For a good drawing D of G we can do the following: cut the sphere along the edges of S(w). This results in a disk bounded by two copies of each edge of S(w) and $n-1$ copies of v (this is Kynčl’s star-cut representation [11]). Isomorphically redraw the disk, so that its boundary is a convex polygon. Since the drawing D was good, each edge of G crosses each boundary edge at most once, so an edge of G consists of at most n arcs in the star-cut drawing. Replacing each crossing with a degree-4 dummy vertex gives us a plane graph. Triangulating that drawing yields a 3-connected plane graph with a unique embedding (up to orientation-preserving homeomorphisms) by Whitney’s theorem [13]. We can now apply Tutte’s spring embedding theorem [18] to find a straight-line embedding of the plane graph in which the bounding convex polygon remains the same. Removing the edges we added, and converting dummy vertices back into crossings, gives us a drawing isomorphic to D in which every arc is a polygonal-arc consisting of at most $n^2$ straight-line segments (since each arc can cross at most $\left( {\begin{array}{c}n\\ 2\end{array}}\right) <n^2$ other arcs). Since each edge of G consists of at most n arcs, each edge of G consists of at most $n^3$ straight-line segments.

We now build star-cuts drawings for both $D_1$ and $D_2$ using the same convex (geometric) polygon based on S(w). Slightly perturbing points inside the disk, we can ensure there is no overlap between edges or crossing points. In this simultaneous drawing, there are at most $n^6$ crossings between any two edges (independently of whether they are based on $D_1$ or $D_2$).

We now glue the pieces of S(w) back together, to obtain a drawing D which contains a drawing $D'_1$ isomorphic to $D_1$ and a drawing $D^*_2$ isomorphic to $D_2$, and in this joint drawing S(w) is drawn the same way. In particular, all vertices are in the same location. Since $D_1$ and $D'_1$ are isomorphic, we can apply a homeomorphism of the plane to $D=D'_1\cup D^*_2$ that turns $D'_1$ into $D_1$. The result of that homeomorphism on $D^*_2$ is the drawing $D'_2$ we need for (i) and (ii).

Claim (iii) is achieved by homeomorphisms: in a small neighborhood of each vertex, rotate the edges of $D'_2$ until they satisfy (iii). This will introduce at most one crossing between any two edges incident to the same vertex belonging to different drawings, so the asymptotic analysis is not affected. $\square$

B Good Drawings of $W_4$

Recall that we want to prove Lemma 3.5: the weak isomorphism type of a good drawing of $W_4$ determines the drawing up to isomorphism. The most direct approach would be to study all non-isomorphic drawings of $W_4$ and prove that their weak isomorphism types differ.

There are two obstacles: (i) when listing non-isomorphic drawings, we typically do so for non-labeled graphs, so one has to take into account that there may be non-trivial automorphisms of the drawing (of the unlabeled graph); the drawings in Fig. 8 show that this difference matters; and (ii) there are no reliable lists of non-isomorphic good drawings of $W_4$. There is a such list in the paper by Gronau and Harborth [9], $W_4$ is $G_{31}$ in their list; however, there are at least two obvious mistakes in that list (the 11th drawing of $W_4$ shows a degree-2 vertex, and the 15th drawing looks like it has two degree-4 vertices). This is probably due to copying mistakes, but makes this a less than ideal basis for a proof. Hence, we opt for a direct proof.

Remark B.1

It would be highly desirable to have a tool that is able to generate all good drawings (maybe even bad [5] drawings) of small graphs (labeled and unlabeled) automatically, in some verifiable manner. There are algorithms in the literature (going as far back as Eggleton [6]), but implementations tend to focus on drawings of small complete, or small complete bipartite graphs.

Proof of Lemma 3.5

Fix a good drawing of $W_4$. The drawing determines a weak isomorphism type. We have to show that we can reconstruct the drawing, up to isomorphism, from the weak isomorphism type. It is sufficient to determine the rotation at each vertex, and the order and direction in which edges cross each other (for this, we fix an arbitrary orientation of the edges). Let q be the unique vertex of degree 4. Edges incident to q are called spokes, the remaining edges are cycle edges.

Claim 1: The rotation at q is determined. Let uv and xy be two disjoint cycle edges. Then the two 3-cycles quv and qxy intersect in q; the underlying curves of those two cycles may touch or cross in q: they touch, if edges qu, uv, qv cross edges qx, xy, qy an even number of times, otherwise they cross, and we can tell which, based on the weak isomorphism type. This test tells us whether qu and qv are neighbors in the rotation at q or not. If they are not neighbors, the rotation at q has to be uxvy (possibly requiring an orientation-reversing homeomorphism). Otherwise, there are two possible rotations: uvxy or uvyx (again up to isomorphism). We can tell which is the case, by checking whether qu and qx are neighbors (using the other two triangles at q).

Claim 2: The rotation at all cycle vertices is determined. Fix a cycle vertex u; for one of the two cycle edges uv incident to u, edges qu and qv must be neighbors in the rotation at q (they cannot both be opposite to qu in the rotation at q). Up to isomorphism, there are then four possible good drawings of the four spokes together with uv, as shown in Fig. 13.

The weak isomorphism type tells us which of the four cases we are in. Vertex u is incident to two cycle edges: uv, and a second edge e, which is either ux or uy. In all four cases, we know whether the other endpoint of e lies inside the triangle quv or not, and the weak isomorphism type tells us how often e crosses edges qu, uv, qv. From this, we can determine if the end of e at u is inside or outside the triangle, determining the rotation at u.

Claim 3: For every pair of edges, the direction of crossing is determined. Only pairs of independent edges can cross, so consider a pair of independent edges which cross. Suppose one of the two edges is a spoke qx, then the other edge must be a cycle edge uv. Consider the triangle quv. We know (using rotation at q and weak isomorphism type), whether x lies inside or outside of quv, and we know whether qx starts inside or outside quv. This tells us in which direction qx crosses uv. Otherwise, we have two cycle edges uv and xy which cross. Applying an orientation-preserving homeomorphism if necessary, we can choose the direction of crossing in this case. Since a good drawing of a $C_4$ can have at most one crossing, this happens at most once.

Claim 4: The order of crossing along each edge is determined. We argue the cases of spokes and cycle edges separately.

Consider a spoke qx (with at least two crossings, otherwise the order of crossings is trivial). In a good drawing of a $W_4$ a spoke can only cross two edges, the two cycle edges uv and vy (with $x\ne y,v,y$), so qx must cross both. The triangle quv splits the sphere into two regions, and we (arbitrarily) call the region containing the end of qx at q the inside of quv; so x will lie on the outside of quv, as shown in the left illustration of Fig. 14. If vy starts inside quv (something we can tell using the rotation at v), then qx crosses vy before it crosses uv. Otherwise, vy starts outside of quv. In this case, qx can cross uv and vy in either order, but the order depends on the direction of vy crossings qx, something we already know. Hence, we can determine the order.

We are left with the case of a cycle edge, call it xy; again, we assume it has at least two crossings. Let quv be the triangle on the remaining two spokes. In a good drawing of the $W_4$ the cycle edge xy can only cross edges qu, qv, and uv, and it must cross at least two of them (otherwise the order is trivial). In this case, we think of x as lying on the outside of quv, and xy crosses the boundary two or three times. In the case of two crossings, the order of crossings along xy is directly determined by the direction of the crossings (going inside quv first, then leaving again, second). In the case of three crossings, there are two orders consistent with the directions; see the two illustrations in the middle and on the right in Fig. 14; the vertices are relabeled $\{a,b,c\}=\{q,u,v\}$ to reduce the number of cases. Now y is incident to two of $\{a,b,c\}$, so it must be incident to a or b. Let us assume that there is an edge ya (the case yb is the same). If the edge ay starts inside the triangle abc, we are in the left case, otherwise, we are in the right case (since ay cannot cross xy). Since we know the rotation at a (by Claim 3), we can decide which case we are in, and determine the order of crossings along xy. $\square$

Fig. 13

The four good drawings of triangle quv with all four spokes

Fig. 14

Left: triangle quv with spoke qx crossing uv; vy can start inside or outside. Middle, right: triangle quv (labels renamed abc) and edge xy; directions of crossing of xy with ab, bc, and ac are the same in both drawings, the order of crossing is not

Corollary B.2

For $G\in \{K_5, K_5-K_2\}$, the weak isomorphism type of a good drawing of G determines the drawing up to isomorphism.

Proof

Fix a good drawing of G. Suppose an edge f is crossed by edges g and h (which the weak isomorphism type tells us). Then g and h must have a shared endpoint (since G contains no three independent edges). Then $\{e,g,h\}$ are part of a $W_4$-subgraph of G, and by Lemma 3.5 the order and direction of crossing of g and h with f is determined. Using this, we can find the order of all edges crossing f. $\square$