Let G be a graph, and x, y are two vertices of G. x and y are called twins of G if
xy ∈E(G) andNG(x) =NG(y). Twins emerge when we duplicate vertices. For any vertices
v ∈V(G), by duplicating a vertex v, we mean that we add a new vertex w to G and let w
be adjacent to all neighbors of v and v itself. Clearly, v and w are twins in a new graph.
If G is a claw-free graph, then duplicating any vertex of G results in a claw-free graph. A 3-connected claw-free graph is called twins-free if every pair of twins appears only in the
3-cut. Recall that a Halin graph H = T ∪C is a plane graph. Following the definition of
H, we notice that for any stem xofT, there aredegT(x) many facesFx containing xand an edge onC. We call such a face an x-face.
Let G be a graph. A spanning Halin subgraph H = T ∪C of G is called a strong
spanning Halin subgraph of G if for any stem x∈V(T), there is an x-face Fx such that x is adjacent (in G) to two end vertices x1, x2 of the edge of Fx on C. For any two different stems x and x∗, we have x
1x2 6=x∗1x∗2 if x and x∗ are not adjacent in T.
Note that the induced subgraph on all stems of T is still a tree, denoted by T′. Every leaf ofT′ is adjacent to at least two consecutive vertices of V(C) in T. So to check whether
H is a strong spanning Halin subgraph, we only need to investigate that for every stem x of
T′, there is a x-face F
x that contains an edge uv of C such that both u and v are adjacent
tox in G.
Lemma 4.2.1. Let G be a graph and {x, y}be a pair of twins of G. IfG\ {y} has a strong
spanning Halin subgraph, then G also has one.
Proof. Let H = T ∪C be a strong spanning Halin subgraph of G\ {y}. We will
V(C) or x is a stem ofT. Case 1: Assume that x∈V(C).
Let x+ and x− be the successor and the processor of x on C. Set v as the stem of
T is adjacent to x. After replacing x by x, y, we insert y in C to obtain a cycle C∗ =
(C\ {x+x, xx−})∪ {x+x, xy, yx−}and a tree T∗ =T ∪ {vy}.
We are now checking that H∗ = T∗ ∪C∗ is a strong spanning Halin subgraph of G.
For any stem w ∈V(T), if w =v, then w has two neighbors x, y inG∗ and xy is on C∗; if
w 6= v, then NG(w) contains two consecutive vertices u and u+. If uu+ 6= x−x, then u, u+
are also consecutive on C∗; if uu+ =xx−, then y, x− are two consecutive neighbors of won
C∗. In any case, we are done.
Case 2: Suppose that x is a stem of T.
In this case, we insert y between x1, x2 by letting C∗ = (C\ {x1x2})∪ {x1y, x2y} and
a tree T∗ =T ∪ {xy}. Clearly, x
1, y ∈ NG(x) are two consecutive vertices on C∗. For each
stemz withz 6=x, let z1, z2 ∈NG(z) be two associated neighbors ofz, which are consecutive
on C. If z1z2 = x1x2, which implies x and z are adjacent in T, then yz ∈ E(G) and y, x2
are two consecutive vertices associate with z on C∗. If z1z2 =6 x1x2, then z1, z2 are still two
consecutive vertices associate with z onC∗. Clearly, each stem is associated with a distinct
edge of C∗.
In Chapter 12, we will show that ifGis 3-connected {claw, Z3}-free or {claw, B1,2}-free
and admits a twins in its 3-cut, then G contains a spanning Halin subgraph. Thus, from
the next chapter on (except Chapter 12), we always assume G is twins-free and try to find
a strong spanning Halin subgraph, instead of a spanning Halin subgraph, in G.
4.3 One special case of 3-connected {claw, Z3}-free graph
In this section, we assume G is a 3-connected {claw, Z3}-free graph with|V1| ≥2, and
at least one of V1 and V2 is not a clique. Then the following proposition is showing that G
Proposition 6. Let G be an n-vertex 3-connected {claw, Z3}-free graph with |V1| ≥ 2, and
at least one of V1 and V2 is not a clique, then G contains a strong spanning Halin subgraph.
Proof. Let G be a graph satisfying the condition of Proposition 6. We will find a
strong spanning Halin subgraph through the following sequence of claims.
Claim 4.3.1. For any x∈S, the induced subgraph hV2∪ {x}i does not contain an induced
path xw1w2w3 and |N2(x)| ≥2.
Proof: To prove the first part of the claim, suppose on the contrary that there is an
x ∈ S and an induced path xw1w2w3 with w1, w2, w3 ∈ V2. Since |V1| ≥ 2, there are two
vertices u1, u2 ∈N1(x), then hu1, u2, x;w1w2w3i is an induced Z3, a contradiction.
We now prove the second part of the claim. If there exist x∈S such that |N2(x)|= 1,
let N2(x) = {w}. By the above statement, we know V2 = {w} ∪N22(x). Since G is claw-
free, N2(w)\ {v} is a clique, as well as V2. This in turn implies V1 is not a clique. Let w′, w′′ ∈N2
2(x). Since N1(x) is a clique, V1\N1(x)6=∅. Let xu1u3 be an induced path with u1, u3 ∈V1. Then hw′, w′′, w;xu1u3iis an induced Z3, giving a contradiction.
For any vertexx∈S, since|V1| ≥2, we have|N1(x)| ≥2. Following Claim 3.1, we have |N2(x)| ≥ 2. So |Ni(x)| ≥ 2 for both i ∈ [1,2]. Thus V1 and V2 are symmetric if we only
use the property |Ni(x)| ≥2. We assume, without loss of generality, thatV2 is not a clique.
Moreover, in the following proof, we letxbe a vertex inSand W1, W2,· · ·, Wkbe the vertex sets of components of hN2
2(x)i. Let Wi′ =N(Wi) for each i ∈ [1, k]. Clearly Wi′ ⊆ N2(x) ,
otherwise, Z3 will be found.
Claim 4.3.2. Wi∪Wi′ is a clique for all i∈[1, k].
Proof: For anyw∈W′
i, we haveN(w)⊇Wi, otherwise there is an induced pathxww1w2
with w1, w2 ∈ W, giving a contradiction to Claim 4.3.1. Then E(Wi, wi′) contains every possible edge between Wi and Wi′. Since G is claw-free, Wi is a clique. Since Wi′ ∈N2(w),
Claim 4.3.3. If N1(x)6=V1, then N22(x) is an independent set. Thus |W1| =|W2|=· · ·= |Wk|= 1.
Proof: SinceV1\N1(x)6=∅, there exists an induced pathxu1u3 withu1, u3 ∈V1. If there
exists i∈[1, k] such that|Wi| ≥2, denotew1, w2 ∈Wi andw′ ∈Wi′, then hw1, w2, w′;xu1u3i
is a Z3, giving a contradiction.
Claim 4.3.4. For any y∈S\ {x}, we have
1) N2(x) =N2(y) and xy ∈E(G) if N2(y)∩N22(x) =∅.
2) N2
2(x) =W1 and N2(y) =W1∪W1′ if N2(y)∩N22(x)6=∅.
Proof 1) Suppose that N2(y)∩N22(x) = ∅. Since V2 is not a clique, N22(x)6= ∅. Thus,
there exists an induced pathxw1w2 with w1, w2 ∈V2. We can show thatN(y)∩N(w2)6=∅.
Otherwise, let w ∈ N(y)\N(w2). Then yww1w2 is an induced P4. This path and two
neighbors of y in V1 induce a Z3, showing a contradiction. Similarly, we can prove that
N(w2)⊆N(y). We now claim that N2(x)\N(w2)⊆N(y). Otherwise, there exists a vertex w′′ ∈N(x)\N(y), then hw
1;y, w2, w′′i is a claw.
2) Suppose that N2(y)∩N22(x) 6= ∅. Recall that N22(x) is a union of disjoint cliques W1, W2,· · · , Wk. We assume, without loss of generality,N2(y)∩W1 6=∅. Letw1 ∈W1∩N(y)
and w′
1 ∈N(w1)∩N(x). Since N2(y) is a clique,N2(y)⊆W1∪W1′. IfN22(x)\W1 6=∅, then
N2
2(x)\W1 *N2(y)∪N22(y), which contradicts Claim 4.3.1. ThusN22(x) =W1. We now show
thatN2(y) = W1∪W1′. SinceV2 is not a clique, there exists a vertexw′ ∈N2(x)\W1∪W1′. If
there exists a vertex w′′∈W′
1\N(y), then there is an induced path yw1w′′w′, together with
the two neighbors ofyinV1, it will induce aZ3. ThusW1′ ⊆N2(y). For anyw′′′ ∈W1\ {w1},
since w′′′, y, w′ ∈N(w′
1) and Gis claw-free, w′′′y∈E(G), which implies W1 ⊆N2(y).
Following Claim 4.3.4, S could be divided into two subsets, S1 and S2, such that S1 = {y∈S|N2(y) = N2(x), xy ∈E(G)} and S2 ={y∈S|N2(y) =W1∪W1′}.
Claim 4.3.5. Both S1∪N2(x) and S2∪W1 ∪W1′ are cliques.
exists w1 ∈ W1′ and w2 ∈ N2(x)\ {W1′}. For any y1, y2 ∈ S2, to avoid hw1;y1, y2, w2i be a
claw, y1y2 ∈E(G). Together with Claim 4.3.1, we have S2∪W1∪W1′ as a clique.
Claim 4.3.6. If S2 6=∅, then |W1′| ≥2.
Proof: Since V2 is not a clique, N2(x)\W1′ is a component of G\(S1 ∪W1′). By the
minimality of |S|, we have |S1 ∪ W1′| ≥ |S| ≥ |S1 ∪S2|, which implies |W1′| ≥ |S2|. If |W′
1|= 1, then |S2| = 1, which implies S2∪W1′ is a vertex cut of size 2, contradicting that
G is 3-connected.
Since V1 and V2 are symmetric in this case, we can get similar properties for hV1i and
can partition S into two subsets: S′
1 = {y∈ S|N1(y) = N1(x)} and S2′ = {y∈ S|N1(y) = U1∪U1′}, where U1 =N12(x) and U1′ =N(U1)∩N(x).
Our goal is to find a strong spanning Halin subgraph H in G. First we want to find
the part of H inhV2∪Siby considering two cases.
Case 1: S2 =∅.
In this case, N2(y) = N2(x) for any y ∈ S. Since G is 3-connected, |Wi′| ≥ 3. Let
w′ i1, w ′ i2, w ′ i3 ∈ W ′ i, V2′ = N2(x)\(∪ki=1{wi′1, w ′ i2, w ′ i3}) = {w ′ 1, w2′,· · · , w′t′}, Pi =wi′2Piw ′ i3 be hamiltonian paths in (Wi∪Wi′)\ {wi′2}fori∈[1, k] andP0 =w1P0wt′ be a hamiltonian path inhV′
2i. LetS ={x1, x2,· · ·, xt}andP′ =x3P′xtbe a hamiltonian path inS\ {x1, x2}. Let C2 =P0P1· · ·PkP′∪ {x2w1}be a path and the vertices on the star {x1} ∪(∪ik=1{wi′2}), where
{x1} is the center, be stems ofT2 with NC(x1) =V(S)∪V2′ and NC(w′i2) =Wi ∪ {w ′ i1, w
′ i3} for all i∈[1, k]. Then T2 is a HIST of hV2∪Si and V(C2) ={u∈T2|degT2(u) = 1}.
Case 2: S2 6=∅.
In this case, bothS1∪N2(x) andS2∪W1∪W1′ are cliques andN22(x) =W1. We letS1 = {x1, x2,· · · , xt1}and S2 ={y1, y2,· · · , yt2}, where we assumet1 ≥2 since|S| ≥3. In partic- ular, we regard y1 is the same vertex as xt in Case 1. We denoteW1 ={w21, w22,· · · , w2k2},
N2(x) = {w11, w12,· · ·, w1k} and assume w11, w12 ∈ W ′ 1 by Claim 4.3.6. Let P′ = x2P′xt1, P′′ = y t2P ′′y
N2(x)\ {w11} and W1, respectively. Let C2 =P ′P
1P2P′′ be a path and {x1, w11} be stems of T2, with NC2(w1) = (V(P
′)\ {x
2})∪V(P1)∪V(P2)∪V(P′′) and NC2(x1) ={x2}. Then
T2 is a HIST of hV2∪Si and V(C2) ={u∈T2|degT2(u) = 1}.
We now find the other part of H inhV1∪ {x1}i by considering three following cases.
Case 1: We assume that|V1| ≥3 and it is not a clique.
If |V1| ≥ 3 and V1 is not a clique, similarly as V2 is not a clique, we can also find a
HIST T1 in hV1 ∪ {x1}i and a path C1 goes through all leaves ofT1. Moreover, we can also
guarantee C1 ∪ T1 is a planar graph and x1 is one stem of T1. We may denote the two
endpoints of C1 by u1 and u2.
Case 2: We assume that|V1| ≥3 and it is a clique.
Let u1, u2, u3 ∈ V1 such that u3x1, u1x2, u2xt ∈E(G). We denote by C1 =u1C1u2 as a
hamiltonian path in V1\ {u3}and T1 as a HIST ofhV1∪ {x1}iwith stemu1, which connects
to all leaves.
Case 3: We assume that|V1| ≤2.
Note that N1(x) = V1 for all x ∈ S. Let C1 = u1u2 and T1 be a HIST of hV1 ∪ {x1}i
with stem x1, which connects to leaves {u1, u2}.
Note that we can assume u1x2, u2y1 ∈ E(G). This is clearly true if V1 is clique or S′
2 = ∅ or S2 = ∅. If neither S2 nor S2′ is empty, since x2 is an arbitrary vertex of S1 and y1 is an arbitrary vertex of S2, we only need to show that x2 ∈ S1′ and y1 ∈ S2′. Suppose
this is not true. Then either x2, y1 ∈S1′, which implies S =S1′, or x2, y1 ∈S2′, which implies S2∪(S1 \ {x1})⊆S2′. Contradicting to the assumption that S2′ 6=∅ and |S1′| ≥2.
Now we letT =T1∪T2,C =C1∪C2∪ {u1x2, u2y1}and H =T ∪C. Then T is a HIST
of G and C is a cycle obtained by connecting all leaves of T, H is planar. Moreover, the
subgraph induced on all stems ofT is a star with centerx. Furthermore, we have|S−1| ≥2 or x1w12 ∈ E(G). Thus, H is a strong spanning Halin subgraph of G. We only give the
x1 x2 xt1 y1 yt2 w11 w12 w13 w21 w2k 2 u′ 11=u1 u′ 12 u′ 13 u′ 21 u′ 22 u′ 23 u11 u21 ut′=u2