
4 Buchin, Evans, Frati, Kostitsyna, L¨offler, Ophelders, Wolff
and contains the line `uv through uand v. Let Γ0be the image of Γunder
a clockwise rotation around `uv by 90◦. Note that Γ0is contained in Π0. Now
let Γ00 be the image of Γ0under another clockwise rotation around `uv by 90◦.
Note that Γ00 is a flipped copy of Γand is contained in Π. Consider the linear
morphs hΓ, Γ 0iand hΓ0, Γ 00 i. In each of them, every vertex travels on a line that
makes a 45◦-angle with both Πand Π0, and all these lines are parallel. Due
to the linearity of the morph and the fact that both pre-image and image are
planar, all vertices stay coplanar during both linear morphs (although, unlike in a
true rotation, the intermediate drawing size changes continuously). In particular,
every intermediate drawing is crossing-free, and uand v(as well as all the points
on `uv) are fixed points.
Operation 2: Outer face change. Let Gbe a biconnected plane graph, let Γ
be a planar straight-line drawing of G, and let fbe a face of Γ.
Lemma 2. There exists a 4-step 3D crossing-free morph from Γto a planar
straight-line drawing Γ000 of Gwhose embedding is the same as the one of Γ,
except that the outer face of Γ000 is f.
We implement Operation 2, which proves Lemma 2, using the stereographic
projection. Let Πbe the plane z= 0, which contains Γ. Let Sbe a sphere that
contains Γin its interior and is centered on a point in the interior of f. Let Γ0be
the 3D straight-line drawing obtained by projecting the vertices of Gfrom their
positions in Γvertically to the Northern hemisphere of S. Let Γ00 be determined
by projecting the vertices of Γ0centrally from the North Pole of Sto Π. Both
projections define linear morphs: hΓ, Γ 0iand hΓ0, Γ 00 i. Indeed, any intermediate
drawing is crossing-free since the rays along which we project are parallel in
hΓ, Γ 0iand diverge in hΓ0, Γ 00 i, and there is a one-to-one correspondence between
the points in the pre-image and in the image. Since the morph also inverts the
rotation system of Γ00 with respect to Γ, we apply Operation 1 to Γ00 , which,
within two morphing steps, flips Γ00 and yields our final drawing Γ000 .
Operation 3: Component flip. Let Gbe a biconnected plane graph, and let
{u, v}be a split pair of G. Let G1, . . . , Gkbe the split components of Gwith
respect to {u, v}. Let Γbe a planar straight-line drawing of Gin which uand
vare incident to the outer face, as in Fig. 2a. Relabel G1, . . . , Gkso that they
appear in clockwise order G1, . . . , Gkaround u, where G1and Gkare incident
to the outer face of Γ. Let iand jbe two (not necessarily distinct) indices with
1≤i≤j≤kand with the following property1: If Gcontains the edge (u, v),
then this edge is one of the components Gi, . . . , Gj. Operation 3 allows us to flip
the embedding of the components Gi, . . . , Gj(and to incidentally reverse their
order), while leaving the embedding of the other components of Gunchanged.
This is formalized in the following.
Lemma 3. There exists an O(n)-step 3D crossing-free morph from Γto a pla-
nar straight-line drawing Γ0of Gin which the embedding of G`is the flip of the
1This is a point where our operations differ from the ones of Angelini et al. [4]. Indeed,
their flip operation applies to any sequence of components of G, while ours does not.