For simplicity, we denote by S1 =V20 ={x1, x2,· · · , xm1}={w01, w02,· · · , w0t0}.
Our goal is to find a strong spanning Halin subgraphH inG. We divide this processor
into two steps. In the first step, we find the part of H in hV2 ∪S1i and in the second step,
Step 1: Finding a spanning subgraph H2 of H in hV2∪S1i
Case 1: Assume S2 =∅.
We let Q = w0t0w1t1· · ·wl−1tl
−1 be a path. If for any j ∈ [0, l−1], we have |V2j| 6= 4. Since for any i ∈ [0, l − 1], both V2j \ {wj2, wjtj−2, wjtj} and V2l are cliques, there exist hamiltonian paths, say P2j = wj1P2jwjtj−2 and P2l = wl1P2lwl2, in them, respectively. Let
P2 =w02w0t0−1w12· · ·w(l−1)tl−1−1 and C2 =P20P21· · ·P2tP2 and all vertices of Q be stems of
T2 withNC2(wjtj) =V(P2j)∪ {wj2, wjsj−1}for allj ∈[0, l−2] andNC2(w(l−1)t(l
−1)) ={w|w∈
V2l−1 ∪V2l}. Then T2 is a HIST of hV2 ∪S1i and V(C2) ={w∈V2∪S1|degT2(w) = 1}. Let
H2 =T2 ∪C2, then H2 is planar (See Figure 6.1).
w11 w12 w1t1 w1t 1−1 w1t 1−2w21 w22 w2t 2 w2t 2−1 w2t 2−2 w31 w32 x2 x1 xm1 xm1−1 xm1−2
Figure 6.1. S2 =∅ and |V2j| 6= 4 for all j ∈[1, l−1].
If there exists j ∈ [0, l−1] such that |V2j| = 4, first we find T2 and C2 in hV2 ∪S1i as
above, then apply Swap-Operation to the HIST (See Figure 6.2).
Swap-Operation:
1) Swapping the positions of w(j+1)1 and w(j+1)2 on the pathC2. 2) Putting wj3 adjacent towj1 and w(j+1)2 along the path C2. 3) Keeping all other vertices’ positions are the same.
After performing Swap-Operation, we get a newT2 and C2, then the new T2 is a HIST
ofhV2∪S1iandV(C2) ={w∈V2∪S1|degT2(w) = 1}. LetH2 =T2∪C2, then H2 is planar. Case 2: Assume S2 6=∅.
w24 w11 w11 w12 w12 w1t 1 w1t 1 w1t 1−1 w1t 1−1 w1t 1−2 w1t 1−2 w21 w21 w22 w22 w23 w2t 2 w2t 2−1 w2t 2−2 w3t 3−1 w3t 3−1 w3t 3−2 w3t 3−2 w3t 3 w3t 3 w31 w31 w32 w32 |V22| ≥5 |V22|= 4
Figure 6.2. An example of swap operation with |V22|= 4 and|V22| ≥5.
Since for all j ∈ [0, l − 1], both V2j \ {wjtj} and V2t are cliques, there exist hamil- tonian paths, say P2j = wj1P2jwjtj−1 and P2l = wl1P2lwl2, in them, respectively. Let
C2 = P20P21· · ·P2l and all vertices of Q be stems in T2 with NC2(wjtj) = V(P2j} for all
j ∈[1, l−2] and NC2(w(l−1)t(l
−1)) ={w|w∈V2t−1∪V2t}. Then T2 is a HIST ofhV2∪S1i and
V(C2) ={w∈V2∪S1|degT2(w) = 1}. LetH2 =T2∪C2, then H2 is planar (See Figure 6.3).
w11 w1t 1 w1t 1−1w21 w2t 2 w2t 2−1 w3t 3 w31 x1 xm1 xm1−1 Figure 6.3. S2 6=∅.
Step 2: Finding a spanning subgraph H1 of H in hV1∪S2∪ {xt}i
Case 1: Assume S2 =∅.
If k ≥ 2 and |V1i| 6= 4 for any i ∈ [1, k − 1]. Since V11 \ {v11, v1s1−2, v1s1}, V1i \
{vi2, visi−1, visi}, for any i ∈ [2, k −1], and V1k are cliques, there exist hamiltonian path- s, say P11 = v12P11v1s1−1, P1i = vi1P1ivisi−2 and P1k = vk1P1kvk2, in them, respective- ly. Let P1 = v22v2t2−1v32· · ·v(k−1)vt(k−1)−1 and Q
′ = v
(k−1)sk
P11P1P1k· · ·P13P12 ∪ {v21v1t1−2} and all vertices of Q
′ be stems of T
1 with NC(v11) =
V(P11), NC(v1s1) = {v1s1−2}, NC(visi) = V(P1i)∪ {vi2, visi−1}, for all i ∈ [2, k − 2], and
NC(v(k−1)sl
k−1) ={v|v ∈V1k∪V1k−1}. Then T1 is a HIST of hV1∪ {xt}i and V(C1) = {v ∈
V1|degT2(v) = 1}. Let H1 =T1∪C1, then H1 is planar.
If k ≥ 2 and there exists i ∈ [1, k−1] such that |V1j| = 4, then we can apply Swap- Operation tohV1i as tohV2i. Similarly, we can find a HIST T1 and a pathC1 inhV1∪ {xt}i, then H1 =T1∪C1 is planar (See Figure 6.4).
v11 v12 v1s1 v1s 1−1 v1s 1−2 v21 v22 v23 v2s 2−1 v31 v32 v41 v42 v3s3 v3s 3−2 v24 xm1
Figure 6.4. S2 =∅,k = 4 and|V12|= 4 and |V13| 6= 4.
If k = 1, which means N1(x1) = V11 = V1 is a clique and |V1| ≥ 3. Then there exists
a hamiltonian path, say C1 = v12C1v1n, in hV1 \ {v11}i. Let {v11, xt} be stems of T1 with
NC1(v11) = {v|v ∈ V1 \ {v11}}. Then T1 is a HIST of hV1 ∪ {xt}i and V(C1) = {v ∈
T1|degT2(v) = 1}. Let H1 =T1∪C1, then H1 is planar.
If k = 1 and V1 = {v11, v12}(or V1 = {v11}). By Claim 4.1.2, N1(xi) = {v1, v2} for all
xi ∈ S1. Let C1 = v11v12 and v11xt, v12xt ∈ E(T1). Then T1 is a HIST of hV1 ∪ {xt}i and
V(C1) ={v ∈V1|degT2(v) = 1}. Let H1 =T1∪C1, then H1 is planar. Case 2: Suppose that S2 6=∅. Note that vn1, vn1−1 ∈N1(y) for ally ∈S2.
Case 2.1: Assume that k ≥2.
Case 2.1.1: Assume that |V11| ≥4.
If |V1k| ≥ 2, since V11\ {v11, v1s1}, V1i \ {visi}, for all i ∈ [2, k], and S2 are cliques, there exist hamiltonian paths, say P11 = v12P11v1s1−1, P1i = vi1P1ivisi−1 and P
′ = y
in them, respectively. Set C1 = P11P12· · ·P1sP′. Let Q′ = vkskv(k−1)sk
−1· · ·v1s1v11xt and all vertices of Q′ be stems of T
1 with NC1(v11) = V(P11)\ {v1s1−1}, NC(v1s1) = {v1s1−1},
NC(visi) = V(P1i), for all i ∈ [2, k−1], NC(vksk) = {v|v ∈ V1k∪S2}. Then T1 is a HIST of hV1∪S2∪ {xt}i and V(C1) = {v ∈ V1∪S2|degT2(v) = 1}. Let H1 =T1∪C1, then H1 is planar (See Figure 6.5).
v11 v12 v1s1 v1s 1−1 v21 v2s2 v2s 2−1 v31 v3s 3−1 v3s 3−1 y1 xm1 ym2 Figure 6.5. S2 =∅ and k = 3.
If |V1k| = 1, then v(k−1)sk−1y, vksky ∈ E(G) for all y ∈ S2 and vkskv(k−1)s(k
−1)−1 ∈ E(G) since |S1| ≥ 2. Similarly as above, we can define P11, P12, · · ·, P1(k−1) and P′. Let P1k =
v(k−1)s(k
−1)−1vksky1 and C1 = P11P12· · ·P1sP
′. Set Q′ = v
(k−1)sk
−1v(k−2)sk−2 · · ·v1s1v11xt be stems ofT1 withNC1(v11) =V(P11)\ {v1s1−1},NC(v1s1) ={v1s1−1},NC(visi) =V(P1i) for all
i∈[2, k−2] andNC(v(k−1)sk
−1) =V(P1k−1)∪S2∪{vksk}. ThenT1is a HIST ofhV1∪S2∪{xt}i and V(C1) ={v ∈V1∪S2|degT2(v) = 1}. Let H1 =T1∪C1, then H1 is planar.
Case 2.1.2: Suppose that |V11|= 3.
If |V12| ≥ 3, we delete the edge v2t2v21 from E(T) and add the edge v1t1v21 to E(T). Similarly as Case 2.1.1, we will find a HIST T1 and a path C1 in hV1∪S2∪ {xt}i.
If |V12|= 2, then V11 = {v11, v12, v13, v21, v22}. From Claim 4.1.2, we know v22y, v21y ∈
E(G) for all y ∈ S2 and v12xt ∈ E(G). In additional, v12v21, v13v22 ∈ E(G). Let C1 =
v11v13v22∪P
′andQ′ =v
21v12xtbe paths and all vertices ofQ
′ be stems ofT
1withNC1(v12) =
{v11, v13} and NC1(v21) = {v22} ∪ {y|y ∈ S2}. Then T1 is a HIST of hV1 ∪S2 ∪ {xt}i and
If|V12|= 1, then V11={v11, v12, v13, v21}. Since |S1| ≥2 and V1 is minimum subject to
S being minimum, we have v21v11, v21v12, v21v13 ∈ E(G), which implies V1 is a clique. Let
C1 ={v12v21} ∪P
′ and {v
13, v11, xt} be stems of T1 with NC1(v11) = {v12} and NC1(v13) =
{v21}∪{y|y∈S2}. ThenT1is a HIST ofhV1∪S2∪{xt}iandV(C1) ={v ∈V1∪S2|degT2(v) = 1}. Let H1 =T1 ∪C1, then H1 is planar.
Case 2.2: Suppose that k = 1, which means N1(x1) = V11 = V1 is a clique. We still let
P′ =y
1P′ym
If|V1| ≥4, let P1 =v2P1vn1−1 be a hamiltonian path inhV1\ {v1, vn1}i andC1 =P1P ′. Set {vn1, v1, xt} be stems of T1 with NC1(vn1) = {vn1−1} ∪ V(P
′) and N
C1(v11) = {v|v ∈
V(P1)\ {vn1−1}}. Then T1 is a HIST ofhV1∪S2∪ {xt}iandV(C1) ={v ∈V1∪S2|degT2(v) = 1}. Let H1 =T1 ∪C1, then H1 is planar.
If |V1| = 3(similarly as |V1| = 2). Let C1 = {v11v13} ∪ P ′ and {v 12, xt} be stems of T1 with NC1(v12) = {v11, v13} ∪ V(P ′). Then T 1 is a HIST of hV1 ∪S2 ∪ {xt}i and
V(C1) ={v ∈V1∪S2|degT2(v) = 1}. Let H1 =T1∪C1, then H1 is planar.
Thus, let T =T1∪T2, C =C1∪C2 and H =T ∪C, thenH is a strong spanning Halin
subgraph in G.
If |V1| = 1, denote by V1 = {v}. Since N2(y) 6= ∅, we have wn2y ∈ E(G) for all
y ∈ S2. Moreover |N2(ym2)| ≥ m2 subject to S being minimum. Let vxt ∈ E(G) and
C = C2P′ ∪y1vx1, add vxt to E(T2) and add wn2(or wn2−1 if |V2l| = 1) to the stem set of
T2. Then T = T2 is a HIST of G with NC(wn2) = {y|y ∈ S2} ∪ {wn2−1}. Let H = T ∪C,
then H is a strong spanning Halin subgraph in G.
We now let T =T1∪T2 and C=C1∪C2(note thatT1 =∅ and C1 =∅ when |V1|= 1).