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Some results on the total proper k-connection number

Some results on the total proper k-connection number 1IntroductionIn this paper, all graphs under our consideration are simple, finite and undirected. We follow the notation and terminology of [1]. For a graph GG, we denote by V(G),E(G)V\left(G),E\left(G)and diam(G){\rm{diam}}\left(G)the vertex set, edge set and diameter of GG, respectively. The distance between two vertices uuand vvin a connected graph GG, denoted by dist(u,v){\rm{dist}}\left(u,v), is the length of a shortest path between them in GG. The eccentricity of a vertex vvin GGis defined as eccG(v)=maxx∈V(G)dist(v,x){{\rm{ecc}}}_{G}\left(v)={\max }_{x\in V\left(G)}{\rm{dist}}\left(v,x). For convenience, a set of internally pairwise vertex disjoint paths will be called disjoint.In recent years, colored notions of connectivity in graphs becomes a new and active subject in graph theory. Stating from rainbow connection [2], rainbow vertex connection [3] and total rainbow connection [4,5] appeared later. Many researchers are working in this field, and a lot of papers have been published in journals, see [6,7,8, 9,10,11, 12,13,14, 15,16] for details. The reader can also see [17] for a survey, [18] for a dynamic survey and [19] for a new monograph on this topic.In 2012, Borozan et al. [20] introduced the concept of proper kk-connection number. A path in an edge-colored graph is a proper path if any two adjacent edges on the path differ in color. An edge-colored graph is proper k-connected if any two distinct vertices of the graph are connected by kkdisjoint proper paths. The proper k-connection number of a kk-connected graph GG, denoted by pck(G)p{c}_{k}\left(G), is defined as the smallest number of colors that are needed in order to make GGproper kk-connected. For more results, the reader can see [21,22, 23,24] for details.As a natural generalization, Jiang et al. [25] presented the concept of proper vertex kk-connection number. A path in a vertex-colored graph is a vertex proper path if any two internal adjacent vertices of the path differ in color. A vertex-colored graph is proper vertex k-connected if any two distinct vertices of the graph are connected by kkdisjoint vertex proper paths. For a kk-connected graph GG, the proper vertex k-connection number of GG, denoted by pvck(G)pv{c}_{k}\left(G), is defined as the smallest number of colors required to make GGproper vertex kk-connected.Motivated by the concept of total chromatic number of graph, now for proper connection and proper vertex connection, the concept of total proper connection was introduced by Jiang et al. [26]. A total coloring of a graph GGis a mapping from the set V(G)∪E(G)V\left(G)\cup E\left(G)to some finite set of colors. A path in a total-colored graph is a total proper path if the coloring of the edges and internal vertices is proper, that is, any two adjacent or incident elements of edges and internal vertices on the path differ in color. A total-colored graph is total proper k-connected if any two distinct vertices of the graph are connected by kkdisjoint total proper paths. For a connected graph GG, the total proper k-connection number of a kk-connected graph GG, denoted by tpck(G){{\rm{tpc}}}_{k}\left(G), is defined as the smallest number of colors that are needed in order to make GGtotal proper kk-connected. For convenience, we write tpc(G){\rm{tpc}}\left(G)for tpc1(G){{\rm{tpc}}}_{1}\left(G). Obviously, tpc(G)≤tpc2(G)≤tpc3(G){\rm{tpc}}\left(G)\le {{\rm{tpc}}}_{2}\left(G)\le {{\rm{tpc}}}_{3}\left(G). By [26], if GGis complete, then tpc(G)=1{\rm{tpc}}\left(G)=1; if GGhas a Hamiltonian path that is not complete, then tpc(G)=3{\rm{tpc}}\left(G)=3. Note that if GGis a nontrivial connected graph and HHis a connected spanning subgraph of GG, then tpc(G)≤tpc(H){\rm{tpc}}\left(G)\le {\rm{tpc}}\left(H).In this paper, we investigate the total proper connection number of a graph GGunder some constraints on its complement graph G¯\overline{G}.Theorem 1.1Let GGbe a connected graph of order n≥3n\ge 3, if diam(G¯){\rm{diam}}\left(\overline{G})does not belong to the following two cases: (i) diam(G¯)=2,3{\rm{diam}}\left(\overline{G})=2,3, (ii) G¯\overline{G}contain exactly two components and one of them is trivial, then tpc(G)≤4{\rm{tpc}}\left(G)\le 4.For the remaining cases, tpc(G){\rm{tpc}}\left(G)can be very large as discussed in Section 2. Then we add a constraint, i.e., we let G¯\overline{G}be triangle-free. Hence, GGis claw-free, and we can derive our next main result:Theorem 1.2For a connected graph GG, if G¯\overline{G}is triangle-free, then tpc(G)=3{\rm{tpc}}\left(G)=3.Recall that a clique of a graph is a set of mutually adjacent vertices, and that a maximum clique is a clique of the largest possible size in a given graph. The clique number ω(G)\omega \left(G)of a graph GGis the number of vertices in a maximum clique in GG. Let GGbe a connected graph, and let XXbe a maximum clique of GG. We say that NX(u){N}_{X}\left(u)is the set of neighbors of uuin G[X]G\left[X]and dX(u)=∣NX(u)∣{d}_{X}\left(u)=| {N}_{X}\left(u)| . Let F=G[V(G)\X]F=G\left[V\left(G)\backslash X]. Kemnitz and Schiermeyer [9] considered graphs with rc(G)=2rc\left(G)=2and large clique number. In this paper, we characterize graphs with small total proper connection number with respect to their large clique number. If ω(G)=n\omega \left(G)=n, then GGis a complete graph, which implies tpc(G)=1{\rm{tpc}}\left(G)=1. If GGis connected and ω(G)=n−1\omega \left(G)=n-1, then GGhas a Hamiltonian path, and so tpc(G)=3{\rm{tpc}}\left(G)=3. For the cases ω(G)=n−2,n−3\omega \left(G)=n-2,n-3, we obtain the following three main results.Figure 1All connected cubic graphs of order 8.Theorem 1.3Let GGbe a connected graph of order nn. If ω(G)=n−2\omega \left(G)=n-2and XXis a maximum clique of GGwith V(G)\X={u1,u2}V\left(G)\backslash X\right=\left\{{u}_{1},{u}_{2}\right\}, then tpc(G)=3{\rm{tpc}}\left(G)=3.Theorem 1.4Let GGbe a connected graph of order nn, diam(G)=2{\rm{diam}}\left(G)=2. If ω(G)=n−3\omega \left(G)=n-3and XXis a maximum clique of GGwith V(G)\X={u1,u2,u3}V\left(G)\backslash X\right=\left\{{u}_{1},{u}_{2},{u}_{3}\right\}, then tpc(G)=3{\rm{tpc}}\left(G)=3or tpc(G)=4{\rm{tpc}}\left(G)=4for the following case F≅3K3,∣NX(u1)∩NX(u2)∩NX(u3)∣=1F\cong 3{K}_{3},| {N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{2})\cap {N}_{X}\left({u}_{3})| =1and dX(u1)=dX(u2)=dX(u3)=1{d}_{X}\left({u}_{1})={d}_{X}\left({u}_{2})={d}_{X}\left({u}_{3})=1.Theorem 1.5Let GGbe a connected graph of order nn, diam(G)≥3{\rm{diam}}\left(G)\ge 3. If ω(G)=n−3\omega \left(G)=n-3and XXis a maximum clique of GGwith V(G)\X={u1,u2,u3}V\left(G)\backslash X\right=\left\{{u}_{1},{u}_{2},{u}_{3}\right\}, then tpc(G)=3{\rm{tpc}}\left(G)=3, or tpc(G)=4{\rm{tpc}}\left(G)=4and one of the following holds. (i)F≅3K3,NX(u)∩NX(v)≠∅F\cong 3{K}_{3},{N}_{X}\left(u)\cap {N}_{X}\left(v)\ne \varnothing and dX(u)=dX(v)=1{d}_{X}\left(u)={d}_{X}\left(v)=1, where uuand vvare any two distinct vertices in V(G)\XV\left(G)\backslash X.(ii)F≅3K3,dX(u1)=dX(u2)=dX(u3)=1F\cong 3{K}_{3},{d}_{X}\left({u}_{1})={d}_{X}\left({u}_{2})={d}_{X}\left({u}_{3})=1and for any two vertices in V(G)\XV\left(G)\backslash X, there is no common neighbor in G[X]G\left[X].For an integer n≥3n\ge 3, the circular ladder CL2n{{\rm{CL}}}_{2n}of order 2n2nis a cubic graph constructed by taking two copies of the cycle Cn{C}_{n}on disjoint vertex sets (u1,u2,…,un)\left({u}_{1},{u}_{2},\ldots ,{u}_{n})and (v1,v2,…,vn)\left({v}_{1},{v}_{2},\ldots ,{v}_{n}), then joining the corresponding vertices uivi{u}_{i}{v}_{i}for 1≤i≤n1\le i\le n. The Möbius ladder M2n{M}_{2n}of order 2n2nis obtained from the ladder by deleting the edges u1un{u}_{1}{u}_{n}and v1vn{v}_{1}{v}_{n}, and then inserting two edges u1vn{u}_{1}{v}_{n}and unv1{u}_{n}{v}_{1}. Subscripts are considered modulo nn, and we can derive our next main result:Theorem 1.6Let nnbe an integer with n≥3n\ge 3. Then(i)tpc(CL2n)=tpc2(CL2n)=3{\rm{tpc}}\left({{\rm{CL}}}_{2n})={{\rm{tpc}}}_{2}\left({{\rm{CL}}}_{2n})=3, tpc3(CL2n)=4{{\rm{tpc}}}_{3}\left({{\rm{CL}}}_{2n})=4.(ii)tpc(M2n)=tpc2(M2n)=3{\rm{tpc}}\left({M}_{2n})={{\rm{tpc}}}_{2}\left({M}_{2n})=3, tpc3(M2n)=4{{\rm{tpc}}}_{3}\left({M}_{2n})=4.In [7], Fujie-Okamoto et al. investigated the rainbow kk-connection numbers of all small cubic graphs of order 8 or less. In this paper, we determine the total proper kk-connection numbers of all small cubic graphs of order 8 or less. We can easily verify that all such cubic graphs have orders 4, 6, or 8, and those with orders 4 or 6 are K4,K3,3{K}_{4},{K}_{3,3}, and K3□K2{K}_{3}\hspace{0.33em}\square \hspace{0.33em}{K}_{2}(where □\square denotes Cartesian product). In [27], it was shown that all connected cubic graphs of order 8 are Q3,M8{Q}_{3},{M}_{8}, F1,F2{F}_{1},{F}_{2}, and F3{F}_{3}, and these graphs are depicted in Figure 1. Our last main result is stated as follows:Theorem 1.7(i)tpc(K4)=1{\rm{tpc}}\left({K}_{4})=1, tpc2(K4)=3{{\rm{tpc}}}_{2}\left({K}_{4})=3, tpc3(K4)=4{{\rm{tpc}}}_{3}\left({K}_{4})=4.(ii)tpc(K3,3)=tpc2(K3,3)=3{\rm{tpc}}\left({K}_{3,3})={{\rm{tpc}}}_{2}\left({K}_{3,3})=3, tpc3(K3,3)=4{{\rm{tpc}}}_{3}\left({K}_{3,3})=4.(iii)tpc(K3□K2)=tpc2(K3□K2)=3{\rm{tpc}}\left({K}_{3}\hspace{0.33em}\square \hspace{0.33em}{K}_{2})={{\rm{tpc}}}_{2}\left({K}_{3}\hspace{0.33em}\square \hspace{0.33em}{K}_{2})=3, tpc3(K3□K2)=4{{\rm{tpc}}}_{3}\left({K}_{3}\hspace{0.33em}\square \hspace{0.33em}{K}_{2})=4.(iv)tpc(Q3)=tpc2(Q3)=3{\rm{tpc}}\left({Q}_{3})={{\rm{tpc}}}_{2}\left({Q}_{3})=3, tpc3(Q3)=4{{\rm{tpc}}}_{3}\left({Q}_{3})=4.(v)tpc(M8)=tpc2(M8)=3{\rm{tpc}}\left({M}_{8})={{\rm{tpc}}}_{2}\left({M}_{8})=3, tpc3(M8)=4{{\rm{tpc}}}_{3}\left({M}_{8})=4.(vi)tpc(F1)=tpc2(F1)=3{\rm{tpc}}\left({F}_{1})={{\rm{tpc}}}_{2}\left({F}_{1})=3, tpc3(F1)=4{{\rm{tpc}}}_{3}\left({F}_{1})=4.(vii)tpc(F2)=tpc2(F2)=3{\rm{tpc}}\left({F}_{2})={{\rm{tpc}}}_{2}\left({F}_{2})=3.(viii)tpc(F3)=tpc2(F3)=3{\rm{tpc}}\left({F}_{3})={{\rm{tpc}}}_{2}\left({F}_{3})=3, tpc3(F3)=4{{\rm{tpc}}}_{3}\left({F}_{3})=4.2Proof of Theorem 1.1In order to prove Theorem 1.1, we need the following lemmas.Lemma 2.1[26] For 2≤m≤n2\le m\le n, we have tpc(Km,n)=3{\rm{tpc}}\left({K}_{m,n})=3.Lemma 2.2[26] If GGis a complete multipartite graph that is neither a complete graph nor a tree, then tpc(G)=3{\rm{tpc}}\left(G)=3.Let NG¯i(x)={v:distG¯(x,v)=i}{N}_{\overline{G}}^{i}\left(x)=\left\{v:{{\rm{dist}}}_{\overline{G}}\left(x,v)=i\right\}, where 0≤i≤30\le i\le 3, and NG¯4(x)={v:dist(x,v)≥4}{N}_{\overline{G}}^{4}\left(x)=\left\{v:{\rm{dist}}\left(x,v)\ge 4\right\}. In this paper, we use NG¯i{N}_{\overline{G}}^{i}instead of NG¯i(x){N}_{\overline{G}}^{i}\left(x)for convenience. Then NG¯0={x}{N}_{\overline{G}}^{0}=\left\{x\right\}and NG¯1=NG¯(x){N}_{\overline{G}}^{1}={N}_{\overline{G}}\left(x).Lemma 2.3For a connected graph GG, if G¯\overline{G}is connected and diam(G¯)≥4{\rm{diam}}\left(\overline{G})\ge 4, then tpc(G)≤4{\rm{tpc}}\left(G)\le 4.ProofChoose a vertex xxwith eccG¯(x)=diam(G¯){{\rm{ecc}}}_{\overline{G}}\left(x)={\rm{diam}}\left(\overline{G}). By the definition of NG¯i{N}_{\overline{G}}^{i}, we know uv∈E(G)uv\in E\left(G)for any u∈NG¯i,v∈NG¯ju\in {N}_{\overline{G}}^{i},v\in {N}_{\overline{G}}^{j}with ∣i−j∣≥2| i-j| \ge 2. Now we define a total coloring of GGas follows: assign color 1 to the edges xuxufor u∈NG¯2u\in {N}_{\overline{G}}^{2}, all edges between NG¯1{N}_{\overline{G}}^{1}and NG¯3{N}_{\overline{G}}^{3}, and all vertices and edges in NG¯4{N}_{\overline{G}}^{4}; assign color 2 to the edges xuxufor u∈NG¯3u\in {N}_{\overline{G}}^{3}, all edges between NG¯2{N}_{\overline{G}}^{2}and NG¯4{N}_{\overline{G}}^{4}, and all vertices and edges in NG¯1{N}_{\overline{G}}^{1}; assign color 3 to the edges xuxufor u∈NG¯4u\in {N}_{\overline{G}}^{4}, all edges between NG¯1{N}_{\overline{G}}^{1}and NG¯4{N}_{\overline{G}}^{4}, and all vertices and edges in NG¯2,NG¯3{N}_{\overline{G}}^{2},{N}_{\overline{G}}^{3}; assign color 4 to the vertex xx.We prove that there is a total proper path between any two vertices uuand vvof GG. It is trivial when uv∈E(G)uv\in E\left(G). Thus, we only need to consider the pairs u,v∈NG¯iu,v\in {N}_{\overline{G}}^{i}or u∈NG¯i,v∈NG¯i+1u\in {N}_{\overline{G}}^{i},v\in {N}_{\overline{G}}^{i+1}. Note that P=xx3x1x4x2P=x{x}_{3}{x}_{1}{x}_{4}{x}_{2}is a total proper path, where xi∈NG¯i{x}_{i}\in {N}_{\overline{G}}^{i}. By means of the path PP, we can find that uuand vvare connected by some total proper path for any u∈NG¯i,v∈NG¯i+1u\in {N}_{\overline{G}}^{i},v\in {N}_{\overline{G}}^{i+1}. If i=1i=1, then P=ux3xx2x4vP=u{x}_{3}x{x}_{2}{x}_{4}vis a total proper path, where xi∈NG¯i{x}_{i}\in {N}_{\overline{G}}^{i}. If i=2i=2, then P=uxx4vP=ux{x}_{4}vis a total proper path, where x4∈NG¯4{x}_{4}\in {N}_{\overline{G}}^{4}. If i=3i=3, then P=ux1x4x2xvP=u{x}_{1}{x}_{4}{x}_{2}xvis a total proper path, where xi∈NG¯i{x}_{i}\in {N}_{\overline{G}}^{i}. If i=4i=4, then P=uxx2vP=ux{x}_{2}vis a total proper path, where x2∈NG¯2{x}_{2}\in {N}_{\overline{G}}^{2}. Hence, tpc(G)≤4{\rm{tpc}}\left(G)\le 4.□Proof of Theorem 1.1Assume that G¯\overline{G}is connected. Since diam(G¯)≥4{\rm{diam}}\left(\overline{G})\ge 4, we have tpc(G)≤4{\rm{tpc}}\left(G)\le 4by Lemma 2.3. Assume that G¯\overline{G}is disconnected. By the assumption, we know that there exist either at least three connected components or exactly two nontrivial components. Let Gi¯\overline{{G}_{i}}be the components of G¯\overline{G}with ti=∣V(Gi¯)∣{t}_{i}=| V\left(\overline{{G}_{i}})| , where 1≤i≤h1\le i\le h. Then GGcontains a connected spanning subgraph Kt1,t2,…,th{K}_{{t}_{1},{t}_{2},\ldots ,{t}_{h}}, and we have tpc(G)≤tpc(Kt1,t2,…,th)=3{\rm{tpc}}\left(G)\le {\rm{tpc}}\left({K}_{{t}_{1},{t}_{2},\ldots ,{t}_{h}})=3by Lemma 2.2. Note that GGis not complete. Thus, tpc(G)≥3{\rm{tpc}}\left(G)\ge 3, and so tpc(G)=3{\rm{tpc}}\left(G)=3.□Next, we will give three examples to show that tpc(G){\rm{tpc}}\left(G)can be arbitrarily large if one of the following three conditions holds: diam(G¯)=2,diam(G¯)=3,G¯{\rm{diam}}\left(\overline{G})=2,{\rm{diam}}\left(\overline{G})=3,\overline{G}contains exactly two connected components and one of them is trivial.Figure 2The graph of Example 2.4.Example 2.4For the graph G¯\overline{G}shown in Figure 2, we choose a vertex xxwith eccG¯(x)=diam(G¯){{\rm{ecc}}}_{\overline{G}}\left(x)={\rm{diam}}\left(\overline{G}). Let NG¯1(x)={ui∣1≤i≤k}{N}_{\overline{G}}^{1}\left(x)=\left\{{u}_{i}| 1\le i\le k\right\}, NG¯2(x)={vj∣1≤j≤k}{N}_{\overline{G}}^{2}\left(x)=\left\{{v}_{j}| 1\le j\le k\right\}, and let E(G¯)={xui∣1≤i≤k}∪{ui1ui2∣1≤i1,i2≤k}∪{vj1vj2∣1≤j1,j2≤k}∪{uivj∣1≤i,j≤k}\{uivi∣1≤i≤k}E\left(\overline{G})=\left\{x{u}_{i}| 1\le i\le k\right\}\cup \left\{{u}_{{i}_{1}}{u}_{{i}_{2}}| 1\le {i}_{1},{i}_{2}\le k\right\}\cup \left\{{v}_{{j}_{1}}{v}_{{j}_{2}}| 1\le {j}_{1},{j}_{2}\le k\right\}\left\cup \left\{{u}_{i}{v}_{j}| 1\le i,j\le k\right\}\backslash \left\{{u}_{i}{v}_{i}| 1\le i\le k\right\}, where k≥3k\ge 3. Obviously, diam(G¯)=2{\rm{diam}}\left(\overline{G})=2and GGis a tree. Then tpc(G)=Δ(G)+1=k+1{\rm{tpc}}\left(G)=\Delta \left(G)+1=k+1by [26, Theorem 1]. Observe that tpc(G){\rm{tpc}}\left(G)will be arbitrarily large based on the increase of kk.Figure 3The graph of Example 2.5.Example 2.5For the graph G¯\overline{G}shown in Figure 3, we choose a vertex xxwith eccG¯(x)=diam(G¯){{\rm{ecc}}}_{\overline{G}}\left(x)={\rm{diam}}\left(\overline{G}). Let NG¯1(x)={ui∣1≤i≤k},NG¯2(x)={vj∣1≤j≤k}{N}_{\overline{G}}^{1}\left(x)=\left\{{u}_{i}| 1\le i\le k\right\},{N}_{\overline{G}}^{2}\left(x)=\left\{{v}_{j}| 1\le j\le k\right\}, and NG¯3(x)={ws∣1≤s≤k}{N}_{\overline{G}}^{3}\left(x)=\left\{{w}_{s}| 1\le s\le k\right\}, where k≥3k\ge 3. Furthermore, let E(G¯)={xui∣1≤i≤k}∪{uivj∣1≤i,j≤k}∪{vjws∣1≤j,s≤k}∪{vj1vj2∣1≤j1,j2≤k}E\left(\overline{G})=\left\{x{u}_{i}| 1\le i\le k\right\}\cup \left\{{u}_{i}{v}_{j}| 1\le i,j\le k\right\}\cup \left\{{v}_{j}{w}_{s}| 1\le j,s\le k\right\}\cup \left\{{v}_{{j}_{1}}{v}_{{j}_{2}}| 1\le {j}_{1},{j}_{2}\le k\right\}. Obviously, diam(G¯)=3{\rm{diam}}\left(\overline{G})=3and GGis a connected graph. Note that NG¯2(x){N}_{\overline{G}}^{2}\left(x)is a stable set in G¯\overline{G}, and each edge between xxand NG¯2(x){N}_{\overline{G}}^{2}\left(x)is a cut edge in GG. Therefore, tpc(G)≥k+1{\rm{tpc}}\left(G)\ge k+1by [26, Proposition 2], and so tpc(G){\rm{tpc}}\left(G)will be arbitrarily large based on the increase of kk.Example 2.6Let G¯\overline{G}contains exactly two components G1¯\overline{{G}_{1}}and G2¯\overline{{G}_{2}}, where G1¯\overline{{G}_{1}}is trivial and G2¯\overline{{G}_{2}}is a clique of G¯\overline{G}. Clearly, GGis a star, and tpc(G)=∣V(G2¯)∣+1{\rm{tpc}}\left(G)=| V\left(\overline{{G}_{2}})| +1. Thus, tpc(G){\rm{tpc}}\left(G)can be made arbitrarily large by increasing ∣V(G2¯)∣| V\left(\overline{{G}_{2}})| .3Proof of Theorem 1.2Lemma 3.1For a connected graph GG, if G¯\overline{G}is connected with diam(G¯)≥4{\rm{diam}}\left(\overline{G})\ge 4, and G¯\overline{G}is triangle-free, then tpc(G)=3{\rm{tpc}}\left(G)=3.ProofChoose a vertex xxwith eccG¯(x)=diam(G¯){{\rm{ecc}}}_{\overline{G}}\left(x)={\rm{diam}}\left(\overline{G}). Since G¯\overline{G}is triangle-free, we know that NG¯1{N}_{\overline{G}}^{1}is a clique in GG. Now we define a total coloring of GGas follows: assign color 1 to the edges xuxufor u∈NG¯2u\in {N}_{\overline{G}}^{2}, all edges between NG¯1{N}_{\overline{G}}^{1}and NG¯3{N}_{\overline{G}}^{3}, and all vertices and edges in NG¯4{N}_{\overline{G}}^{4}; assign color 2 to the edges between NG¯2{N}_{\overline{G}}^{2}and NG¯4{N}_{\overline{G}}^{4}, all vertices and edges in NG¯1{N}_{\overline{G}}^{1}, and the vertex xx; assign color 3 to the edges xuxufor u∈NG¯3,NG¯4u\in {N}_{\overline{G}}^{3},{N}_{\overline{G}}^{4}, all edges between NG¯1{N}_{\overline{G}}^{1}and NG¯4{N}_{\overline{G}}^{4}, and all vertices and edges in NG¯2,NG¯3{N}_{\overline{G}}^{2},{N}_{\overline{G}}^{3}.We prove that there is a total proper path between any two distinct vertices uuand vvin GG. Note that P=xx2x4x1x3P=x{x}_{2}{x}_{4}{x}_{1}{x}_{3}is a total proper path, where xi∈NG¯i{x}_{i}\in {N}_{\overline{G}}^{i}. By means of the path PP, we can find that uuand vvare connected by some total proper path for any u∈NG¯i,v∈NG¯i+1u\in {N}_{\overline{G}}^{i},v\in {N}_{\overline{G}}^{i+1}. Thus, we only need to consider the pairs u,v∈NG¯iu,v\in {N}_{\overline{G}}^{i}. For i=2i=2, P=uxx4vP=ux{x}_{4}vis a total proper path, where x4∈NG¯4{x}_{4}\in {N}_{\overline{G}}^{4}. For i=4i=4, P=uxx2vP=ux{x}_{2}vis a total proper path, where x2∈NG¯2{x}_{2}\in {N}_{\overline{G}}^{2}. For i=3i=3, P=ux1x4x2xvP=u{x}_{1}{x}_{4}{x}_{2}xvis a total proper path, where xi∈NG¯i{x}_{i}\in {N}_{\overline{G}}^{i}. Thus, GGis total proper connected with the above coloring, and so tpc(G)=3{\rm{tpc}}\left(G)=3.□Lemma 3.2Let GGbe a connected graph. If diam(G¯)=3{\rm{diam}}\left(\overline{G})=3and G¯\overline{G}is triangle-free, then tpc(G)=3{\rm{tpc}}\left(G)=3.ProofFor a vertex xxof G¯\overline{G}satisfying eccG¯(x)=diam(G¯)=3{{\rm{ecc}}}_{\overline{G}}\left(x)={\rm{diam}}\left(\overline{G})=3, let ni{n}_{i}represent the number of vertices with distance iifrom xx. If n1=n2=n3=1{n}_{1}={n}_{2}={n}_{3}=1, then G≅P4G\cong {P}_{4}, and so tpc(G)=3{\rm{tpc}}\left(G)=3.Case 1. Two of n1,n2,n3{n}_{1},{n}_{2},{n}_{3}are equal to 1. Without loss of generality, we may assume n1=n2=1{n}_{1}={n}_{2}=1. Since G¯\overline{G}is triangle-free, we have that NG¯3{N}_{\overline{G}}^{3}is a stable set in G¯\overline{G}, and so a clique in GG. We can find that GGhas a Hamiltonian path. Thus, tpc(G)=3{\rm{tpc}}\left(G)=3.Case 2. One of n1,n2,n3{n}_{1},{n}_{2},{n}_{3}is equal to 1. Suppose n2=1{n}_{2}=1. Since G¯\overline{G}is triangle-free, we know that NG¯1{N}_{\overline{G}}^{1}and NG¯3{N}_{\overline{G}}^{3}is a stable set in G¯\overline{G}, and so a clique in GG. Note that GGhas a Hamiltonian path, and so tpc(G)=3{\rm{tpc}}\left(G)=3.Subcase 2.1. n1=1{n}_{1}=1. Since G¯\overline{G}is triangle-free, we obtain that NG¯2{N}_{\overline{G}}^{2}is a clique in GG. Define a total coloring of GGas follows: assign color 3 to the vertex xx, all edges between NG¯2{N}_{\overline{G}}^{2}and NG¯3{N}_{\overline{G}}^{3}, and all edges between NG¯1{N}_{\overline{G}}^{1}and NG¯3{N}_{\overline{G}}^{3}; assign color 2 to the edges xuxufor u∈NG¯2u\in {N}_{\overline{G}}^{2}, and all vertices and edges in NG¯3{N}_{\overline{G}}^{3}; assign color 1 to the edges xuxufor u∈NG¯3u\in {N}_{\overline{G}}^{3}, and all vertices and edges in NG¯1,NG¯2{N}_{\overline{G}}^{1},{N}_{\overline{G}}^{2}. We prove that there is a total proper path between any two distinct vertices uuand vvin GG. Note that P=x1x3xx2P={x}_{1}{x}_{3}x{x}_{2}is a total proper path, where xi∈NG¯i{x}_{i}\in {N}_{\overline{G}}^{i}. By means of the path PP, we know that uuand vvare connected by some total proper path for any u∈NG¯i,v∈NG¯i+1u\in {N}_{\overline{G}}^{i},v\in {N}_{\overline{G}}^{i+1}. For any two vertices u,v∈NG¯3u,v\in {N}_{\overline{G}}^{3}, it is trivial if uv∈E(G)uv\in E\left(G). If uv∉E(G)uv\notin E\left(G), since u,v∈NG¯3u,v\in {N}_{\overline{G}}^{3}, there exist two vertices u′,v′∈NG¯3{u}^{^{\prime} },{v}^{^{\prime} }\in {N}_{\overline{G}}^{3}such that uu′,vv′∈E(G¯)u{u}^{^{\prime} },v{v}^{^{\prime} }\in E\left(\overline{G}). Since G¯\overline{G}is triangle-free, we can see that u′≠v′{u}^{^{\prime} }\ne {v}^{^{\prime} }and vu′,uv′∈E(G)v{u}^{^{\prime} },u{v}^{^{\prime} }\in E\left(G). Then P=uxu′vP=ux{u}^{^{\prime} }via a total proper path. Hence, GGis total proper connected with the above coloring, and so tpc(G)=3{\rm{tpc}}\left(G)=3.Subcase 2.2. n3=1{n}_{3}=1. Since G¯\overline{G}is triangle-free, we know that NG¯1{N}_{\overline{G}}^{1}is a stable set in G¯\overline{G}, and so a clique in GG. Define a total coloring of GGas follows: assign color 3 to the vertex xx, and all edges between NG¯1{N}_{\overline{G}}^{1}and NG¯3{N}_{\overline{G}}^{3}; assign color 2 to the edges xuxufor u∈NG¯2u\in {N}_{\overline{G}}^{2}, and all vertices and edges in NG¯3{N}_{\overline{G}}^{3}; assign color 1 to the edges xuxufor u∈NG¯3u\in {N}_{\overline{G}}^{3}, and all vertices and edges in NG¯1,NG¯2{N}_{\overline{G}}^{1},{N}_{\overline{G}}^{2}. We prove that there is a total proper path between any two distinct vertices uuand vvin GG. Note that P=x1x3xx2P={x}_{1}{x}_{3}x{x}_{2}is a total proper path, where xi∈NG¯i{x}_{i}\in {N}_{\overline{G}}^{i}. By means of the path PP, we obtain that uuand vvare connected by some total proper path for any u∈NG¯i,v∈NG¯i+1u\in {N}_{\overline{G}}^{i},v\in {N}_{\overline{G}}^{i+1}. Let u,vu,vbe any two distinct vertices of NG¯2{N}_{\overline{G}}^{2}, and NG¯3={y}{N}_{\overline{G}}^{3}=\{y\}. If yyis adjacent to any vertex of NG¯2{N}_{\overline{G}}^{2}in G¯\overline{G}, then NG¯2{N}_{\overline{G}}^{2}is a clique in GG, and so GGhas a Hamiltonian path. Otherwise, let Vy{V}_{y}denote the set of neighbors of yyin NG¯2{N}_{\overline{G}}^{2}in GG. We can check that P=uyxvP=uyxvis a total proper path, where u,v∈Vyu,v\in {V}_{y}. If ∣NG¯2\Vy∣=1\left| {N}_{\overline{G}}^{2}\backslash {V}_{y}\right| =1, then P=uyxvP=uyxvis a total proper path, where u∈Vy,v∈NG¯2\Vyu\in {V}_{y},v\left\in {N}_{\overline{G}}^{2}\backslash {V}_{y}. If ∣NG¯2\Vy∣≥2\left| {N}_{\overline{G}}^{2}\backslash {V}_{y}\right| \ge 2, then GGis claw-free since G¯\overline{G}is triangle-free, and G[x∪NG¯2\Vy]G\left[x\left\cup {N}_{\overline{G}}^{2}\backslash {V}_{y}]is a complete graph. Note that P=uyxvP=uyxvis a total proper path, where u∈Vy,v∈NG¯2u\in {V}_{y},v\in {N}_{\overline{G}}^{2}. Thus, GGis total proper connected with the above coloring, and so tpc(G)=3{\rm{tpc}}\left(G)=3.Case 3. n1,n2,n3≥2{n}_{1},{n}_{2},{n}_{3}\ge 2. Since G¯\overline{G}is triangle-free, we have that NG¯1{N}_{\overline{G}}^{1}is a stable set in G¯\overline{G}, and so a clique in GG. If any vertex in NG¯3{N}_{\overline{G}}^{3}is adjacent to all vertices of NG¯2{N}_{\overline{G}}^{2}in G¯\overline{G}, then both NG¯2{N}_{\overline{G}}^{2}and NG¯3{N}_{\overline{G}}^{3}are stable sets in G¯\overline{G}, and so cliques in GG. Thus, GGhas a Hamiltonian path, and so tpc(G)=3{\rm{tpc}}\left(G)=3.Otherwise, we choose a vertex u∈NG¯3u\in {N}_{\overline{G}}^{3}, let Vu{V}_{u}denote the set of neighbors of uuin NG¯2{N}_{\overline{G}}^{2}in GG, we have Vu≠∅,NG¯2{V}_{u}\ne \varnothing ,{N}_{\overline{G}}^{2}. Define a total coloring of GG: assign color 2 to the vertex xx, all vertices and edges in NG¯1{N}_{\overline{G}}^{1}, and all edges between NG¯2,NG¯3{N}_{\overline{G}}^{2},{N}_{\overline{G}}^{3}; assign color 3 to the vertex uu, all edges between NG¯1{N}_{\overline{G}}^{1}and NG¯3\{u}{N}_{\overline{G}}^{3}\backslash \left\{u\right\}, and all edges between xxand Vu{V}_{u}; assign color 2 to the remaining vertices and edges. Note that P=xvux1P=xvu{x}_{1}is a total proper path, where v∈Vu,x1∈NG¯1v\in {V}_{u},{x}_{1}\in {N}_{\overline{G}}^{1}. For any two vertices w,z∈NG¯3w,z\in {N}_{\overline{G}}^{3}, P=wx1uvxzP=w{x}_{1}uvxzis a total proper path, where x1∈NG¯1,v∈Vu{x}_{1}\in {N}_{\overline{G}}^{1},v\in {V}_{u}. For any two vertices w,z∈NG¯2w,z\in {N}_{\overline{G}}^{2}, P=wuxvP=wuxvis a total proper path, where u,v∈Vyu,v\in {V}_{y}. If ∣NG¯2\Vy∣=1\left| {N}_{\overline{G}}^{2}\backslash {V}_{y}\right| =1, then P=uxvP=uxvis a total proper path, where u∈Vy,v∈NG¯2\Vyu\in {V}_{y},v\left\in {N}_{\overline{G}}^{2}\backslash {V}_{y}. If ∣NG¯2\Vy∣≥2\left| {N}_{\overline{G}}^{2}\backslash {V}_{y}\right| \ge 2, since G¯\overline{G}is triangle-free, we know that GGis claw-free, and the subgraph G[x∪NG¯2\Vy]G\left[x\left\cup {N}_{\overline{G}}^{2}\backslash {V}_{y}]is a complete graph. Note that P=uxvP=uxvis a total proper path, where u∈Vy,v∈NG¯2\Vyu\in {V}_{y},v\left\in {N}_{\overline{G}}^{2}\backslash {V}_{y}. For any w∈NG¯2,z∈NG¯3w\in {N}_{\overline{G}}^{2},z\in {N}_{\overline{G}}^{3}, P=wxvux1zP=wxvu{x}_{1}zis a total proper path, where v∈Vy,x1∈NG¯1v\in {V}_{y},{x}_{1}\in {N}_{\overline{G}}^{1}. Similarly, there is a total proper path connecting any two vertices w∈NG¯2,z∈NG¯1w\in {N}_{\overline{G}}^{2},z\in {N}_{\overline{G}}^{1}. Hence, GGis total proper connected, and so tpc(G)=3{\rm{tpc}}\left(G)=3.□Lemma 3.3For a connected graph GG, if G¯\overline{G}is triangle-free and diam(G¯)=2{\rm{diam}}\left(\overline{G})=2, then tpc(G)=3{\rm{tpc}}\left(G)=3.ProofChoose a vertex xxwith eccG¯(x)=diam(G¯)=2{{\rm{ecc}}}_{\overline{G}}\left(x)={\rm{diam}}\left(\overline{G})=2. Since GGis connected, we have n1≥2,n2=1{n}_{1}\ge 2,\hspace{0.33em}{n}_{2}=1or n1,n2≥2{n}_{1},{n}_{2}\ge 2, and there exist two vertices u∈NG¯1,v∈NG¯2u\in {N}_{\overline{G}}^{1},v\in {N}_{\overline{G}}^{2}such that uv∈E(G)uv\in E\left(G). Assume n1≥2{n}_{1}\ge 2and n2=1{n}_{2}=1. Since G¯\overline{G}is triangle-free, we know that NG¯1{N}_{\overline{G}}^{1}is a stable set in G¯\overline{G}, and so a clique in GG. Note that GGhas a Hamiltonian path, and so tpc(G)=3{\rm{tpc}}\left(G)=3.Assume n1,n2≥2{n}_{1},{n}_{2}\ge 2. Observe that NG¯1{N}_{\overline{G}}^{1}is a stable set in G¯\overline{G}since G¯\overline{G}is triangle-free, and so a clique in GG. We show a total coloring of GGas follows: assign color 1 to the vertex xx, the edge uvuvand all vertices in NG¯1\u{N}_{\overline{G}}^{1}\backslash u; assign color 3 to the vertex vvand all edges in NG¯1{N}_{\overline{G}}^{1}; assign color 2 to the remaining vertices and edges. If there exist some vertices w∈NG¯2w\in {N}_{\overline{G}}^{2}with dG¯(w)=n−2{d}_{\overline{G}}\left(w)=n-2, then wwis adjacent to the remaining vertices except xxin G¯\overline{G}. Since diam(G¯)=2{\rm{diam}}\left(\overline{G})=2, there exists an edge w1w2∈E(G¯){w}_{1}{w}_{2}\in E\left(\overline{G})with w1∈NG¯1,w2∈NG¯2{w}_{1}\in {N}_{\overline{G}}^{1},{w}_{2}\in {N}_{\overline{G}}^{2}. Thus, w,w1,w2w,{w}_{1},{w}_{2}is a triangle in G¯\overline{G}, a contradiction. Hence, dG¯(w)<n−2{d}_{\overline{G}}\left(w)\lt n-2for all w∈NG¯2w\in {N}_{\overline{G}}^{2}, and so dG(w)≥2{d}_{G}\left(w)\ge 2. For any z∈NG¯1z\in {N}_{\overline{G}}^{1}, we know that P=xvuzP=xvuzis a total proper path. For any y∈NG¯2\{v}y\left\in {N}_{\overline{G}}^{2}\backslash \left\{v\right\}and z∈NG¯1z\in {N}_{\overline{G}}^{1}, if NG(y)∩NG¯1≠∅{N}_{G}(y)\cap {N}_{\overline{G}}^{1}\ne \varnothing , let w∈NG(y)∩NG¯1w\in {N}_{G}(y)\cap {N}_{\overline{G}}^{1}. Then ywzywzis a total proper path. Otherwise, let NG(y)∩NG¯1=∅{N}_{G}(y)\cap {N}_{\overline{G}}^{1}=\varnothing . We claim that yyis adjacent to all the other vertices of NG¯2{N}_{\overline{G}}^{2}in GG. In fact, for any vertex w∈NG¯2\{y}w\left\in {N}_{\overline{G}}^{2}\backslash \{y\}, there exists a vertex w′∈NG¯1w^{\prime} \in {N}_{\overline{G}}^{1}such that ww′∈E(G¯)w{w}^{^{\prime} }\in E\left(\overline{G}). Since yw′∈E(G¯)y{w}^{^{\prime} }\in E\left(\overline{G}), we know that yw∈E(G)yw\in E\left(G). Then yvuzyvuzis a total proper path. Next we consider w,z∈NG¯2w,z\in {N}_{\overline{G}}^{2}such that wz∉E(G)wz\notin E\left(G). Since G¯\overline{G}is triangle-free, we have that GGis claw-free, and at least one of wwand zzis adjacent to the vv, without loss of generality, assume that wv∈E(G)wv\in E\left(G). Since w,z∈NG¯2w,z\in {N}_{\overline{G}}^{2}, there exist two vertices w′,z′∈NG¯2{w}^{^{\prime} },{z}^{^{\prime} }\in {N}_{\overline{G}}^{2}such that ww′,zz′∈E(G¯)w{w}^{^{\prime} },z{z}^{^{\prime} }\in E\left(\overline{G}), and w′≠z′{w}^{^{\prime} }\ne {z}^{^{\prime} }. Then zw′,wz′∈E(G)z{w}^{^{\prime} },w{z}^{^{\prime} }\in E\left(G)and P=wvuw′zP=wvu{w}^{^{\prime} }zis a total proper path. Thus, GGis total proper connected with the above coloring. Hence, tpc(G)=3{\rm{tpc}}\left(G)=3.□Lemma 3.4Let GGbe a connected graph of order n≥3n\ge 3. If G¯\overline{G}is disconnected and triangle-free, then tpc(G)=3{\rm{tpc}}\left(G)=3.ProofSuppose G¯\overline{G}is triangle-free and contains two connected components one of which is trivial. Let G1¯\overline{{G}_{1}}and G2¯\overline{{G}_{2}}be the two components of G¯\overline{G}, where V(G1¯)={u}V\left(\overline{{G}_{1}})=\left\{u\right\}. Then uuis adjacent to any other vertex in GG. We will consider two cases according to the value of δ\delta , where δ\delta is the minimum degree of GG. If δ=1\delta =1, let d(v)=δd\left(v)=\delta . Since G¯\overline{G}is triangle-free, we know that GGis claw-free, and the subgraph G[V(G)\{v}]G\left[V\left(G)\backslash \left\{v\right\}]is a complete graph. Thus, GGhas a Hamiltonian path, and so tpc(G)=3{\rm{tpc}}\left(G)=3. If δ≥2\delta \ge 2, let d(v)=δ,D=V(G)\{u,v}d\left(v)=\delta ,D\left=V\left(G)\backslash \left\{u,v\right\}, and Vv{V}_{v}be the set of neighbors of vvin GG. Now we define a total coloring of GGas follows: assign color 1 to the vertex vvand all the edges between uuand Vv{V}_{v}; assign color 3 to the vertex uuand all the edges between vvand Vv{V}_{v}; assign color 2 to the remaining vertices and edges. Since GGis claw-free, we can find that the subgraph G[V(G)\{v}∪Vv]G\left[V\left(G)\backslash \left\{v\right\}\right\cup {V}_{v}]is a complete graph, and P=v1uvv2P={v}_{1}uv{v}_{2}is a total proper path, where v1,v2∈Vv{v}_{1},{v}_{2}\in {V}_{v}. For any w∈Vv,z∈D\Vvw\in {V}_{v},z\left\in D\backslash {V}_{v}, we obtain that P=wuzP=wuzis a total proper path. Thus, GGis total proper connected with the above coloring, and so tpc(G)=3{\rm{tpc}}\left(G)=3. Suppose G¯\overline{G}contains at least three connected components or exactly two nontrivial components. Then we have tpc(G)=3{\rm{tpc}}\left(G)=3by the similar proof of Theorem 1.1.□Proof of Theorem 1.2If G¯\overline{G}is connected, the result holds for the case diam(G¯)≥4{\rm{diam}}\left(\overline{G})\ge 4by Lemma 3.1, the case diam(G¯)=3{\rm{diam}}\left(\overline{G})=3by Lemma 3.2, and the case diam(G¯)=2{\rm{diam}}\left(\overline{G})=2by Lemma 3.3. If G¯\overline{G}is disconnected, the result holds by Lemma 3.4.□4Proof of Theorem 1.3Suppose F≅K2F\cong {K}_{2}. Note that GGhas a Hamiltonian path, and thus tpc(G)=3{\rm{tpc}}\left(G)=3. Next, we compute the total proper connection number of GGby proving the following claim.Claim 1. Let GGbe a graph obtained by adding two pendant vertices {u1,u2}\left\{{u}_{1},{u}_{2}\right\}to a vertex v1{v}_{1}of a complete graph Kt{K}_{t}. Then tpc(G)=3{\rm{tpc}}\left(G)=3.ProofSince GGis not a complete graph, we have tpc(G)≥3{\rm{tpc}}\left(G)\ge 3. Now we only need to prove tpc(G)≤3{\rm{tpc}}\left(G)\le 3by the following cases.Case 1. t≡0(mod3)t\equiv 0\left({\rm{mod}}\hspace{0.33em}3). Assign a total coloring ccto GGas follows: Let c(u1v1)=1,c(u2v1)=3c\left({u}_{1}{v}_{1})=1,c\left({u}_{2}{v}_{1})=3; c(v3i+1)=c(v3i+2v3i+3)=2c\left({v}_{3i+1})=c\left({v}_{3i+2}{v}_{3i+3})=2, c(v3i+2)=c(v3i+3v3i+4)=1c\left({v}_{3i+2})=c\left({v}_{3i+3}{v}_{3i+4})=1, and c(v3i+3)=c(v3i+1u3i+2)=3c\left({v}_{3i+3})=c\left({v}_{3i+1}{u}_{3i+2})=3, where 0≤i≤t3−10\le i\le \frac{t}{3}-1. Observe that P1=u1v1v2⋯vt−1vt{P}_{1}={u}_{1}{v}_{1}{v}_{2}\cdots {v}_{t-1}{v}_{t}and P2=u2v1vtvt−1⋯v3v2{P}_{2}={u}_{2}{v}_{1}{v}_{t}{v}_{t-1}\cdots {v}_{3}{v}_{2}are two total proper paths.Case 2. t≡1(mod3)t\equiv 1\left({\rm{mod}}\hspace{0.33em}3). Assign a total coloring ccto GGas follows: Let c(u1v1)=c(vt−1v1)=1,c(vt)=2,c(u2v1)=c(vtv1)=3c\left({u}_{1}{v}_{1})=c\left({v}_{t-1}{v}_{1})=1,c\left({v}_{t})=2,c\left({u}_{2}{v}_{1})=c\left({v}_{t}{v}_{1})=\hspace{-0.08em}3. Let iibe an integer with 0≤i≤t3−10\le i\le &#x230A;\hspace{-0.16em},\frac{t}{3},\hspace{-0.16em}&#x230B;-1, c(v3i+1)=c(v3i+2v3i+3)=2c\left({v}_{3i+1})=c\left({v}_{3i+2}{v}_{3i+3})=2, c(v3i+2)=c(v3i+3v3i+4)=1c\left({v}_{3i+2})=c\left({v}_{3i+3}{v}_{3i+4})=1, and c(v3i+3)=c(v3i+1u3i+2)=3c\left({v}_{3i+3})=c\left({v}_{3i+1}{u}_{3i+2})=3. We can find that P1=u1v1v2⋯vt−1vt{P}_{1}={u}_{1}{v}_{1}{v}_{2}\cdots {v}_{t-1}{v}_{t}and P2=u2v1vt−1⋯v3v2vt{P}_{2}={u}_{2}{v}_{1}{v}_{t-1}\cdots {v}_{3}{v}_{2}{v}_{t}are two total proper paths.Case 3. t≡2(mod3)t\equiv 2\left({\rm{mod}}\hspace{0.33em}3). Assign a total coloring ccto GGas follows: Let c(u1v1)=c(vtv1)=c(vt)=c(vt−2v1)=1,c(vt−1)=2,c(u2v1)=c(vtvt−1)=c(vt−1v2)=3c\left({u}_{1}{v}_{1})=c\left({v}_{t}{v}_{1})=c\left({v}_{t})=c\left({v}_{t-2}{v}_{1})=1,c\left({v}_{t-1})=2,c\left({u}_{2}{v}_{1})=c\left({v}_{t}{v}_{t-1})=c\left({v}_{t-1}{v}_{2})=3. Let iibe an integer with 0≤i≤t3−10\le i\le &#x230A;\hspace{-0.16em},\frac{t}{3},\hspace{-0.16em}&#x230B;-1, c(v3i+1)=c(v3i+2v3i+3)=2c\left({v}_{3i+1})=c\left({v}_{3i+2}{v}_{3i+3})=2, c(v3i+2)=c(v3i+3v3i+4)=1c\left({v}_{3i+2})=c\left({v}_{3i+3}{v}_{3i+4})=1, and c(v3i+3)=c(v3i+1u3i+2)=3c\left({v}_{3i+3})=c\left({v}_{3i+1}{u}_{3i+2})=3. We can easily verify that P1=u1v1v2⋯vt−1vt,P2=u2v1vt−2vt−3⋯v3v2vt−1{P}_{1}={u}_{1}{v}_{1}{v}_{2}\cdots {v}_{t-1}{v}_{t},{P}_{2}={u}_{2}{v}_{1}{v}_{t-2}{v}_{t-3}\cdots {v}_{3}{v}_{2}{v}_{t-1}, and P3=vtv1u2{P}_{3}={v}_{t}{v}_{1}{u}_{2}are three total proper paths.Thus, GGis total proper connected with the above coloring, and so tpc(G)=3{\rm{tpc}}\left(G)=3. This completes the proof of Claim 1.□Suppose F≅2K1F\cong 2{K}_{1}. Assume that NX(u1)∩NX(u2)=∅{N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{2})=\varnothing . Then GGhas a Hamiltonian path, and so tpc(G)=3{\rm{tpc}}\left(G)=3. Otherwise, if ∣NX(u1)∩NX(u2)∣=dX(u1)=dX(u2)=1| {N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{2})| ={d}_{X}\left({u}_{1})={d}_{X}\left({u}_{2})=1, then we know that tpc(G)=3{\rm{tpc}}\left(G)=3from Claim 1. If ∣NX(u1)∩NX(u2)∣≥2| {N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{2})| \ge 2, or ∣NX(u1)∩NX(u2)∣=1| {N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{2})| =1, and max{dX(u1),dX(u2)}≥2{\rm{\max }}\left\{{d}_{X}\left({u}_{1}),{d}_{X}\left({u}_{2})\right\}\ge 2, then we can find that GGhas a Hamiltonian path. Thus, tpc(G)=3{\rm{tpc}}\left(G)=3.5Proof of Theorem 1.4Suppose F≅K3F\cong {K}_{3}or P3{P}_{3}. Note that GGhas a Hamiltonian path, and so tpc(G)=3{\rm{tpc}}\left(G)=3. The following three claims will be used later.Claim 2. Let GGbe a graph obtained by adding a pendant vertex u3{u}_{3}adjacent to vertex u1{u}_{1}or u2{u}_{2}of graph in Claim 1. Then tpc(G)=3{\rm{tpc}}\left(G)=3.ProofWithout loss of generality, assume that u3{u}_{3}is adjacent to u1{u}_{1}. Let c(u1)={1,2,3}\{c(u1v1),c(v1)},c(u1u3)=c(v1)c\left({u}_{1})\left=\left\{1,2,3\right\}\backslash \left\{c\left({u}_{1}{v}_{1}),c\left({v}_{1})\right\}\right,c\left({u}_{1}{u}_{3})=c\left({v}_{1}), and the remaining vertices and edges are assigned the same color as Claim 1. We can verify that GGis total proper connected with the above coloring. Then tpc(G)=3{\rm{tpc}}\left(G)=3. This completes the proof of Claim 2.□Claim 3. Let GGbe a graph obtained by adding three vertices {u1,u2,u3}\left\{{u}_{1},{u}_{2},{u}_{3}\right\}to a complete graph Kt{K}_{t}such that d(u1)=d(u3)=1,d(u2)=2,N(u1)∩N(u3)=∅d\left({u}_{1})=d\left({u}_{3})=1,d\left({u}_{2})=2,N\left({u}_{1})\cap N\left({u}_{3})=\varnothing , and ∣N(u2)∩N(ui)∣=1| N\left({u}_{2})\cap N\left({u}_{i})| =1, where 1≤i≤21\le i\le 2. Then tpc(G)=3{\rm{tpc}}\left(G)=3.ProofWithout loss of generality, assume that N(u2)∩N(u1)=v1,N(u2)∩N(u3)=viN\left({u}_{2})\cap N\left({u}_{1})={v}_{1},N\left({u}_{2})\cap N\left({u}_{3})={v}_{i}. Let c(u3vi)=c(vivi+1),c(v1u2)=c(vivi−1)c\left({u}_{3}{v}_{i})=c\left({v}_{i}{v}_{i+1}),c\left({v}_{1}{u}_{2})=c\left({v}_{i}{v}_{i-1}), and the remaining vertices and edges are assigned the same color as Claim 1. Suppose t≡0(mod3)t\equiv 0\left({\rm{mod}}\hspace{0.33em}3). Note that P1=u1v1v2⋯vt−1vt{P}_{1}={u}_{1}{v}_{1}{v}_{2}\cdots {v}_{t-1}{v}_{t}, P2=u2v1vtvt−1⋯v3v2{P}_{2}={u}_{2}{v}_{1}{v}_{t}{v}_{t-1}\cdots {v}_{3}{v}_{2}, P3=u3viu2{P}_{3}={u}_{3}{v}_{i}{u}_{2}, P4=u1v1v2⋯viu3{P}_{4}={u}_{1}{v}_{1}{v}_{2}\cdots {v}_{i}{u}_{3}, and P5=u3vivi−1⋯v1vtvt−1⋯vi+1{P}_{5}={u}_{3}{v}_{i}{v}_{i-1}\cdots {v}_{1}{v}_{t}{v}_{t-1}\cdots {v}_{i+1}are five total proper paths. Suppose t≡1(mod3)t\equiv 1\left({\rm{mod}}\hspace{0.33em}3). Note that P1=u1v1v2⋯vt−1vt{P}_{1}={u}_{1}{v}_{1}{v}_{2}\cdots {v}_{t-1}{v}_{t}, P2=u2v1vt−1⋯v3v2vt{P}_{2}={u}_{2}{v}_{1}{v}_{t-1}\cdots {v}_{3}{v}_{2}{v}_{t}, P3=u3viu2{P}_{3}={u}_{3}{v}_{i}{u}_{2}, P4=u1v1v2⋯viu3{P}_{4}={u}_{1}{v}_{1}{v}_{2}\cdots {v}_{i}{u}_{3}and P5=u3vivi−1⋯v2vtvt−1⋯vi+1{P}_{5}={u}_{3}{v}_{i}{v}_{i-1}\cdots {v}_{2}{v}_{t}{v}_{t-1}\cdots {v}_{i+1}are five total proper paths. Suppose t≡2(mod3)t\hspace{-0.08em}\equiv \hspace{-0.08em}2\left({\rm{mod}}\hspace{0.25em}3). We can find that P1=u1v1v2⋯vt−1vt{P}_{1}\hspace{-0.08em}=\hspace{-0.08em}{u}_{1}{v}_{1}{v}_{2}\cdots {v}_{t-1}{v}_{t}, P2=u2v1vt−2⋯v3v2vt−1{P}_{2}={u}_{2}{v}_{1}{v}_{t-2}\cdots {v}_{3}{v}_{2}{v}_{t-1}, P3=u3viu2{P}_{3}={u}_{3}{v}_{i}{u}_{2}, P4=u1v1v2⋯viu3{P}_{4}={u}_{1}{v}_{1}{v}_{2}\cdots {v}_{i}{u}_{3}, P5=u3vivi−1⋯v2vt−1vt−2⋯vi+1{P}_{5}={u}_{3}{v}_{i}{v}_{i-1}\cdots {v}_{2}{v}_{t-1}{v}_{t-2}\cdots {v}_{i+1}, and P6=u3vivi−1vt{P}_{6}={u}_{3}{v}_{i}{v}_{i-1}{v}_{t}are six total proper paths. Therefore, GGis total proper connected with the above coloring, and so tpc(G)=3{\rm{tpc}}\left(G)=3. This completes the proof of Claim 3.□Claim 4. Let GGbe a graph obtained by adding a vertex uuto the graph in Claim 1 such that d(u)=2d\left(u)=2and v1{v}_{1}is adjacent to uu. Then tpc(G)=3{\rm{tpc}}\left(G)=3.ProofSince d(u)=2d\left(u)=2, without loss of generality, we assume that vi{v}_{i}is adjacent to uu. Let c(uv1)=1,c(uvi)={1,2,3}\{c(vi),c(vivi−1)}c\left(u{v}_{1})=1,c\left(u{v}_{i})\left=\left\{1,2,3\right\}\backslash \left\{c\left({v}_{i}),c\left({v}_{i}{v}_{i-1})\right\}, and the remaining vertices and edges are assigned the same color as Claim 1. Suppose t≡0(mod3)t\equiv 0\left({\rm{mod}}\hspace{0.33em}3). Note that P1=uvivi−1⋯v1vtvt−1⋯vi+1,P2=uvivi−1⋯v1u1{P}_{1}=u{v}_{i}{v}_{i-1}\cdots {v}_{1}{v}_{t}{v}_{t-1}\cdots {v}_{i+1},{P}_{2}=u{v}_{i}{v}_{i-1}\cdots {v}_{1}{u}_{1}and P3=uv1u2{P}_{3}=u{v}_{1}{u}_{2}are three total proper paths. Suppose t≡1(mod3)t\equiv 1\left({\rm{mod}}\hspace{0.33em}3). Note that P1=uvivi−1⋯v1u1{P}_{1}=u{v}_{i}{v}_{i-1}\cdots {v}_{1}{u}_{1}, P2=uvivi−1⋯v2vtvt−1⋯vi+1{P}_{2}=u{v}_{i}{v}_{i-1}\cdots {v}_{2}{v}_{t}{v}_{t-1}\cdots {v}_{i+1}, and P3=uv1u2{P}_{3}=u{v}_{1}{u}_{2}are three total proper paths. Suppose t≡2(mod3)t\equiv 2\left({\rm{mod}}\hspace{0.33em}3). We can find that P1=uvivi−1⋯v1u1{P}_{1}=u{v}_{i}{v}_{i-1}\cdots {v}_{1}{u}_{1}, P2=uvivi−1⋯v2vt−1vt−2⋯vi+1{P}_{2}=u{v}_{i}{v}_{i-1}\cdots {v}_{2}{v}_{t-1}{v}_{t-2}\cdots {v}_{i+1}, P3=uv1u2{P}_{3}=u{v}_{1}{u}_{2}, and P4=uvivi−1⋯v1vt{P}_{4}=u{v}_{i}{v}_{i-1}\cdots {v}_{1}{v}_{t}are four total proper paths. Hence, GGis total proper connected with the above coloring, and so tpc(G)=3{\rm{tpc}}\left(G)=3. This completes the proof of Claim 4.□Suppose F≅K2+K1F\cong {K}_{2}+{K}_{1}. Let V(K2)={u1,u2}V\left({K}_{2})=\left\{{u}_{1},{u}_{2}\right\}and V(K1)={u3}V\left({K}_{1})=\left\{{u}_{3}\right\}. Since diam(G)=2{\rm{diam}}\left(G)=2, we have NX(u1)∩NX(u2)∩NX(u3)={v}{N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{2})\cap {N}_{X}\left({u}_{3})=\left\{v\right\}, and so tpc(G)=3{\rm{tpc}}\left(G)=3by Claim 2. Suppose F≅3K1F\cong 3{K}_{1}. Assume NX(u1)∩NX(u2)∩NX(u3)=∅{N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{2})\cap {N}_{X}\left({u}_{3})=\varnothing . Then tpc(G)=3{\rm{tpc}}\left(G)=3by Claim 3. Assume NX(u1)∩NX(u2)∩NX(u3)≠∅{N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{2})\cap {N}_{X}\left({u}_{3})\ne \varnothing . If dX(u1)=dX(u2)=dX(u3)=1{d}_{X}\left({u}_{1})={d}_{X}\left({u}_{2})={d}_{X}\left({u}_{3})=1, then tpc(G)≥4{\rm{tpc}}\left(G)\ge 4by [26, Proposition 2]. Define a total coloring of GGas follows: c(u3v)=4c\left({u}_{3}v)=4with v∈NX(u3)v\in {N}_{X}\left({u}_{3}), and the remaining vertices and edges are assigned the same color as Claim 1. We check that any two vertices have a total proper path, and so tpc(G)=4{\rm{tpc}}\left(G)=4. Otherwise, we have dX(u1)+dX(u2)+dX(u3)≥4{d}_{X}\left({u}_{1})+{d}_{X}\left({u}_{2})+{d}_{X}\left({u}_{3})\ge 4. Without loss of generality, let dX(u1)≥2{d}_{X}\left({u}_{1})\ge 2, and u∈X\{v}u\left\in X\backslash \left\{v\right\}where v∈NX(u1)∩NX(u2)∩NX(u3)v\in {N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{2})\cap {N}_{X}\left({u}_{3}). Thus, tpc(G)=3{\rm{tpc}}\left(G)=3by Claim 4.6Proof of Theorem 1.5Case 1. diam(G)=3{\rm{diam}}\left(G)=3. We prove Case 1 by analyzing the structure of FF.Subcase 1.1. F≅K3F\cong {K}_{3}or P3{P}_{3}. Note that GGhas a Hamiltonian path, and so tpc(G)=3{\rm{tpc}}\left(G)=3.Subcase 1.2. F≅K2+K1F\cong {K}_{2}+{K}_{1}. Denote V(K2)={u1,u2}V\left({K}_{2})=\left\{{u}_{1},{u}_{2}\right\}and V(K1)={u3}V\left({K}_{1})=\left\{{u}_{3}\right\}. Suppose NX(u1)∩NX(u3)≠∅{N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{3})\ne \varnothing or NX(u2)∩NX(u3)≠∅{N}_{X}\left({u}_{2})\cap {N}_{X}\left({u}_{3})\ne \varnothing . Without of loss generality, we may assume that NX(u1)∩NX(u3)≠∅{N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{3})\ne \varnothing . Then tpc(G)=3{\rm{tpc}}\left(G)=3by Claim 2. Suppose NX(u1)∩NX(u3)=∅{N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{3})=\varnothing and NX(u2)∩NX(u3)=∅{N}_{X}\left({u}_{2})\cap {N}_{X}\left({u}_{3})=\varnothing . Since diam(G)=3{\rm{diam}}\left(G)=3, we have NX(u1)≠∅{N}_{X}\left({u}_{1})\ne \varnothing and NX(u3)≠∅{N}_{X}\left({u}_{3})\ne \varnothing . Then GGhas a Hamiltonian path, and so tpc(G)=3{\rm{tpc}}\left(G)=3.Subcase 1.3. F≅3K1F\cong 3{K}_{1}. Let V(F)={u1,u2,u3}V\left(F)=\left\{{u}_{1},{u}_{2},{u}_{3}\right\}. Since diam(G)=3{\rm{diam}}\left(G)=3, we have NX(u1)∩NX(u2)∩NX(u3)=∅{N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{2})\cap {N}_{X}\left({u}_{3})=\varnothing . Suppose there exists two vertices ui,uj∈V(F){u}_{i},{u}_{j}\in V\left(F)satisfy NX(ui)∩NX(uj)≠∅{N}_{X}\left({u}_{i})\cap {N}_{X}\left({u}_{j})\ne \varnothing . Without loss of generality, let u1{u}_{1}and u2{u}_{2}satisfy NX(u1)∩NX(u2)≠∅{N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{2})\ne \varnothing and v1∈NX(u1)∩NX(u2){v}_{1}\in {N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{2}). Assume dX(u1)=dX(u2)=1{d}_{X}\left({u}_{1})={d}_{X}\left({u}_{2})=1and vi∈N(u3){v}_{i}\in N\left({u}_{3}). Since GGis not complete, we have tpc(G)≥3{\rm{tpc}}\left(G)\ge 3. To the contrary, suppose there exists a total coloring ccof GGusing three colors such that GGis total proper connected. Since any two vertices of GGare connected by a total proper path, we have c(u1v1)≠c(v1)≠c(u2v1)c\left({u}_{1}{v}_{1})\ne c\left({v}_{1})\ne c\left({u}_{2}{v}_{1}). Without loss of generality, let c(u1v1)=1,c(v1)=2c\left({u}_{1}{v}_{1})=1,c\left({v}_{1})=2and c(u2v1)=3c\left({u}_{2}{v}_{1})=3. Consider the total proper path PPbetween u1{u}_{1}and uu, then the color of vertices and edges in PPfollows the sequence 1,2,3,…,1,2,3,…1,2,3,\ldots ,1,2,3,\ldots . Thus, the value of (c(vi),c(viu))\left(c\left({v}_{i}),c\left({v}_{i}u))is (1,2),(2,3)\left(1,2),\left(2,3), or (3,1)\left(3,1). Consider the total proper path QQbetween u2{u}_{2}and uu, then the color of vertices and edges in QQfollows the sequence 3,2,1,…,3,2,1,…3,2,1,\ldots ,3,2,1,\ldots . But the value of (c(vi),c(viu))\left(c\left({v}_{i}),c\left({v}_{i}u))is (3,2),(2,1)\left(3,2),\left(2,1), or (1,3)\left(1,3), a contradiction. Assign a total coloring ccto GGas follows: c(u1v1)=1,c(v1)=c(viu)=2,c(u2v1)=3c\left({u}_{1}{v}_{1})=1,c\left({v}_{1})=c\left({v}_{i}u)=2,c\left({u}_{2}{v}_{1})=3, assign 4 to the remaining edges, and assign 1 to the remaining vertices. We can check that GGis total proper connected with the above coloring, and so tpc(G)=4{\rm{tpc}}\left(G)=4. Assume dX(u1)+dX(u2)≥3{d}_{X}\left({u}_{1})+{d}_{X}\left({u}_{2})\ge 3, without loss of generality, let dX(u1)≥2{d}_{X}\left({u}_{1})\ge 2. If dX(u3)=1{d}_{X}\left({u}_{3})=1and NX(u1)∩NX(u3)≠∅{N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{3})\ne \varnothing , then we have tpc(G)=3{\rm{tpc}}\left(G)=3by Claim 3; if NX(u1)∩NX(u3)=∅{N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{3})=\varnothing , or NX(u1)∩NX(u3)≠∅{N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{3})\ne \varnothing and dX(u3)≥2{d}_{X}\left({u}_{3})\ge 2, then GGhas a Hamiltonian path, and so tpc(G)=3{\rm{tpc}}\left(G)=3.Now, we may suppose NX(u1)∩NX(u2)=∅,NX(u1)∩NX(u3)=∅{N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{2})=\varnothing ,{N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{3})=\varnothing , and NX(u2)∩NX(u3)=∅{N}_{X}\left({u}_{2})\cap {N}_{X}\left({u}_{3})=\varnothing . Since GGis not complete, we have tpc(G)≥3{\rm{tpc}}\left(G)\ge 3. To the contrary, assume that vi∈N(ui){v}_{i}\in N\left({u}_{i}), and there exists a total coloring ccof GGusing three colors such that GGis total proper connected. Since any two vertices of GGare connected by a total proper path, we have c(u1v1)≠c(v1)c\left({u}_{1}{v}_{1})\ne c\left({v}_{1}). Without loss of generality, let c(u1v1)=1c\left({u}_{1}{v}_{1})=1and c(v1)=2c\left({v}_{1})=2. Consider the total proper path PPbetween u1{u}_{1}and u2{u}_{2}, then the color of vertices and edges in PPfollows the sequence 1,2,3,…,1,2,3,…1,2,3,\ldots ,1,2,3,\ldots . Thus, the value of (c(v2),c(v2u2))\left(c\left({v}_{2}),c\left({v}_{2}{u}_{2}))is (1,2),(2,3)\left(1,2),\left(2,3), or (3,1)\left(3,1). Consider the total proper path QQbetween u1{u}_{1}and u3{u}_{3}, then the color of vertices and edges in QQfollows the sequence 1,2,3,…,1,2,3,…1,2,3,\ldots ,1,2,3,\ldots . Hence, the value of (c(v3),c(v3u3))\left(c\left({v}_{3}),c\left({v}_{3}{u}_{3}))is (1,2),(2,3)\left(1,2),\left(2,3), or (3,1)\left(3,1). Consider the total proper path WWbetween u2{u}_{2}and u3{u}_{3}. If c(v2)=1c\left({v}_{2})=1and c(v2u2)=2c\left({v}_{2}{u}_{2})=2, then the color of vertices and edges in WWfollows the sequence 2,1,3,…,2,1,3,…2,1,3,\ldots ,2,1,3,\ldots . Note that the value of (c(v3),c(v3u3))\left(c\left({v}_{3}),c\left({v}_{3}{u}_{3}))is (2,1),(1,3)\left(2,1),\left(1,3), or (3,2)\left(3,2), a contradiction. If c(v2)=2c\left({v}_{2})=2and c(v2u2)=3c\left({v}_{2}{u}_{2})=3, then the color of vertices and edges in WWfollows the sequence 3,2,1,…,3,2,1,…3,2,1,\ldots ,3,2,1,\ldots . Note that the value of (c(v3),c(v3u3))\left(c\left({v}_{3}),c\left({v}_{3}{u}_{3}))is (2,1),(1,3)\left(2,1),\left(1,3), or (3,2)\left(3,2), a contradiction. If c(v2)=3c\left({v}_{2})=3and c(v2u2)=1c\left({v}_{2}{u}_{2})=1, then the color of vertices and edges in WWfollows the sequence 1,3,2,…,1,3,2,…1,3,2,\ldots ,1,3,2,\ldots . Note that the value of (c(v3),c(v3u3))\left(c\left({v}_{3}),c\left({v}_{3}{u}_{3}))is (2,1),(1,3)\left(2,1),\left(1,3), or (3,2)\left(3,2), a contradiction. Assign a total coloring ccto GGas follows: c(u1v1)=c(v2)=1,c(v1)=c(u2v2)=c(u3v3)=2c\left({u}_{1}{v}_{1})=c\left({v}_{2})=1,c\left({v}_{1})=c\left({u}_{2}{v}_{2})=c\left({u}_{3}{v}_{3})=2, assign 4 to the remaining vertices, and assign 3 to the remaining edges. We can verify that GGis total proper connected with the above coloring, and so tpc(G)=4{\rm{tpc}}\left(G)=4.Case 2. diam(G)≥4.{\rm{diam}}\left(G)\ge 4.Thus, F≅P3F\cong {P}_{3}or F≅K2+K1F\cong {K}_{2}+{K}_{1}. Assume F≅P3F\cong {P}_{3}. Obviously, GGhas a Hamiltonian path, and so tpc(G)=3{\rm{tpc}}\left(G)=3. Assume F≅K2+K1F\cong {K}_{2}+{K}_{1}. Denote V(K2)={u1,u2}V\left({K}_{2})=\left\{{u}_{1},{u}_{2}\right\}and V(K1)={u3}V\left({K}_{1})=\left\{{u}_{3}\right\}, without loss of generality, we have dX(u2)=0,dX(u1)≥1,dX(u3)≥1{d}_{X}\left({u}_{2})=0,{d}_{X}\left({u}_{1})\ge 1,{d}_{X}\left({u}_{3})\ge 1satisfying NX(u1)∩NX(u2)=∅{N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{2})=\varnothing . Hence, we can find that GGhas a Hamiltonian path, and so tpc(G)=3{\rm{tpc}}\left(G)=3.7Proof of Theorem 1.6The proof of Theorem 1.6 follows from the next two lemmas. First, we shall determine the total proper kk-connection numbers of the circular ladders.Lemma 7.1Let nnbe an integer with n≥3n\ge 3. Then tpc(CL2n)=tpc2(CL2n)=3,tpc3(CL2n)=4{\rm{tpc}}\left({{\rm{CL}}}_{2n})={{\rm{tpc}}}_{2}\left({{\rm{CL}}}_{2n})=3,{{\rm{tpc}}}_{3}\left({{\rm{CL}}}_{2n})=4.ProofLet nnbe an integer with n≥3n\ge 3. Since CL2n{{\rm{CL}}}_{2n}contains a Hamiltonian path that is not complete, we have tpc(CL2n)=3{\rm{tpc}}\left({{\rm{CL}}}_{2n})=3. Since tpc2(CL2n)≥tpc(CL2n)=3{{\rm{tpc}}}_{2}\left({{\rm{CL}}}_{2n})\ge {\rm{tpc}}\left({{\rm{CL}}}_{2n})=3, we only need to prove tpc2(CL2n)≤3{{\rm{tpc}}}_{2}\left({{\rm{CL}}}_{2n})\le 3.Case 1. n≡0(mod3)n\equiv 0\left({\rm{mod}}\hspace{0.33em}3). Let n=3tn=3t. Assign a total coloring ccto CL2n{{\rm{CL}}}_{2n}as follows: Let iibe an integer with 0≤i≤t−10\le i\le t\hspace{-0.08em}-\hspace{-0.08em}1, c(u3i+1)=c(u3i+2u3i+3)=c(v3i+2)=c(v3i+3v3i+4)=1c\left({u}_{3i+1})\hspace{-0.08em}=\hspace{-0.08em}c\left({u}_{3i+2}{u}_{3i+3})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{3i+2})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{3i+3}{v}_{3i+4})=1, c(u3i+2)=c(u3i+3u3i+4)=c(v3i+3)=c(v3i+1v3i+2)=2c\left({u}_{3i+2})=c\left({u}_{3i+3}{u}_{3i+4})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{3i+3})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{3i+1}{v}_{3i+2})\hspace{-0.08em}=\hspace{-0.08em}2, and c(u3i+3)=c(u3i+1u3i+2)=c(v3i+1)=c(v3i+2v3i+3)=3c\left({u}_{3i+3})=c\left({u}_{3i+1}{u}_{3i+2})=c\left({v}_{3i+1})=c\left({v}_{3i+2}{v}_{3i+3})=3; c(uivi),c(ui),c(vi)∈{1,2,3}c\left({u}_{i}{v}_{i}),c\left({u}_{i}),c\left({v}_{i})\in \left\{1,2,3\right\}with c(uivi)≠c(ui)≠c(vi)c\left({u}_{i}{v}_{i})\ne c\left({u}_{i})\ne c\left({v}_{i})for 1≤i≤n1\le i\le n. Let xxand yybe any two distinct vertices of CL2n{{\rm{CL}}}_{2n}. By symmetry, we may assume that x=u1x={u}_{1}. If y=uiy={u}_{i}for 2≤i≤n2\le i\le n, then xu2u3⋯ui−1yx{u}_{2}{u}_{3}\cdots {u}_{i-1}yand xunun−1⋯ui+1yx{u}_{n}{u}_{n-1}\cdots {u}_{i+1}yare two total proper paths connecting xxand yy. If y=v1y={v}_{1}, then xyxyand xunun−1⋯u2yx{u}_{n}{u}_{n-1}\cdots {u}_{2}yare two total proper paths connecting xxand yy. If y=viy={v}_{i}, then xv1vnvn−1⋯vi+1yx{v}_{1}{v}_{n}{v}_{n-1}\cdots {v}_{i+1}yand xunun−1⋯uiyx{u}_{n}{u}_{n-1}\cdots {u}_{i}yare two total proper paths connecting xxand yy. Thus, CL2n{{\rm{CL}}}_{2n}is total proper 2-connected with the above coloring.Case 2. n≡1(mod3)n\equiv 1\left({\rm{mod}}\hspace{0.33em}3). Let n=3t+1n=3t+1. Define a total coloring ccof CL2n{{\rm{CL}}}_{2n}as follows: Let iibe an integer with 0≤i≤t−10\le i\le t\hspace{-0.08em}-\hspace{-0.08em}1, c(u3i+1)=c(u3i+2u3i+3)=c(v3i+3)=c(v3i+1v3i+2)=1c\left({u}_{3i+1})\hspace{-0.08em}=\hspace{-0.08em}c\left({u}_{3i+2}{u}_{3i+3})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{3i+3})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{3i+1}{v}_{3i+2})\hspace{-0.08em}=\hspace{-0.08em}1, c(u3i+2)=c(u3i+3u3i+4)=c(v3i+2)=c(v3i+3v3i+4)=3c\left({u}_{3i+2})\hspace{-0.08em}=\hspace{-0.08em}c\left({u}_{3i+3}{u}_{3i+4})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{3i+2})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{3i+3}{v}_{3i+4})=3, and c(u3i+3)=c(u3i+1u3i+2)=c(v3i+1)=c(v3i+2v3i+3)=2c\left({u}_{3i+3})=c\left({u}_{3i+1}{u}_{3i+2})=c\left({v}_{3i+1})=c\left({v}_{3i+2}{v}_{3i+3})=2; c(un)=c(v1vn)=1,c(unvn)=2,c(vn)=c(u1un)=3c\left({u}_{n})=c\left({v}_{1}{v}_{n})=1,c\left({u}_{n}{v}_{n})=2,c\left({v}_{n})=c\left({u}_{1}{u}_{n})=3, and c(ujvj)=3c\left({u}_{j}{v}_{j})=3for 1≤j≤n−11\le j\le n-1. Let xxand yybe any two distinct vertices of CL2n{{\rm{CL}}}_{2n}. We may assume that x=uix={u}_{i}for 1≤i≤n−11\le i\le n-1. If y=ujy={u}_{j}for i≤j≤n−1i\le j\le n-1, then xui+1ui+2⋯uj−1yx{u}_{i+1}{u}_{i+2}\cdots {u}_{j-1}yand xui−1ui−2⋯u1v1v2⋯vn−1un−1un−2⋯uj+1yx{u}_{i-1}{u}_{i-2}\cdots {u}_{1}{v}_{1}{v}_{2}\cdots {v}_{n-1}{u}_{n-1}{u}_{n-2}\cdots {u}_{j+1}yare two total proper paths connecting xxand yy. If y=vjy={v}_{j}for 1≤j≤n−11\le j\le n-1, then xui−1ui−2⋯u1v1v2⋯vj−1yx{u}_{i-1}{u}_{i-2}\cdots {u}_{1}{v}_{1}{v}_{2}\cdots {v}_{j-1}yand xui+1⋯un−1vn−1vn−2⋯vj+1yx{u}_{i+1}\cdots {u}_{n-1}{v}_{n-1}{v}_{n-2}\cdots {v}_{j+1}yare two total proper paths connecting xxand yy. If y=uny={u}_{n}, then xui+1⋯un−1yx{u}_{i+1}\cdots {u}_{n-1}yand xui−1ui−2⋯u1yx{u}_{i-1}{u}_{i-2}\cdots {u}_{1}yare two total proper paths connecting xxand yy. If y=vny={v}_{n}, then xui+1⋯un−1unyx{u}_{i+1}\cdots {u}_{n-1}{u}_{n}yand xui−1ui−2⋯u1v1v2⋯vn−1yx{u}_{i-1}{u}_{i-2}\cdots {u}_{1}{v}_{1}{v}_{2}\cdots {v}_{n-1}yare two total proper paths connecting xxand yy. Assume that x=unx={u}_{n}. If y=vjy={v}_{j}for 1≤j≤n−11\le j\le n-1, then xu1u2⋯un−1vn−1vn−2⋯vj+1yx{u}_{1}{u}_{2}\cdots {u}_{n-1}{v}_{n-1}{v}_{n-2}\cdots {v}_{j+1}yand xunvnv1v2⋯vj−1yx{u}_{n}{v}_{n}{v}_{1}{v}_{2}\cdots {v}_{j-1}yare two total proper paths connecting xxand yy. If y=vny={v}_{n}, then xyxyand xun−1un−2⋯u1v1yx{u}_{n-1}{u}_{n-2}\cdots {u}_{1}{v}_{1}yare two total proper paths connecting xxand yy. Thus, CL2n{{\rm{CL}}}_{2n}is total proper 2-connected with the above coloring.Case 3. n≡2(mod3)n\equiv 2\left({\rm{mod}}\hspace{0.33em}3). Let n=3t+2n=3t+2. Define a total coloring ccof CL2n{{\rm{CL}}}_{2n}as follows: Let iibe an integer with 0≤i≤t−20\le i\le t\hspace{-0.08em}-\hspace{-0.08em}2, c(u3i+1)=c(u3i+2u3i+3)=c(v3i+1)=c(v3i+2v3i+3)=1c\left({u}_{3i+1})\hspace{-0.08em}=\hspace{-0.08em}c\left({u}_{3i+2}{u}_{3i+3})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{3i+1})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{3i+2}{v}_{3i+3})\hspace{-0.08em}=\hspace{-0.08em}1, c(u3i+2)=c(u3i+3u3i+4)=c(v3i+2)=c(v3i+3v3i+4)=3c\left({u}_{3i+2})\hspace{-0.08em}=\hspace{-0.08em}c\left({u}_{3i+3}{u}_{3i+4})=c\left({v}_{3i+2})=c\left({v}_{3i+3}{v}_{3i+4})=3, and c(u3i+3)=c(u3i+1u3i+2)=c(v3i+1)=c(v3i+2v3i+3)=2c\left({u}_{3i+3})\hspace{-0.08em}=\hspace{-0.08em}c\left({u}_{3i+1}{u}_{3i+2})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{3i+1})=c\left({v}_{3i+2}{v}_{3i+3})=2; c(un)=c(un−4)=c(vn−2)=c(vn)=1c\left({u}_{n})=c\left({u}_{n-4})=c\left({v}_{n-2})=c\left({v}_{n})=1, c(un−1)=c\left({u}_{n-1})=c(vn−1)=c(un−3)=c(vn−3)=3c\left({v}_{n-1})\hspace{0.12em}=\hspace{0.12em}c\left({u}_{n-3})\hspace{0.12em}=\hspace{0.12em}c\left({v}_{n-3})\hspace{0.12em}=\hspace{0.12em}3, c(un−2)=c(vn−4)=2c\left({u}_{n-2})\hspace{0.12em}=\hspace{0.12em}c\left({v}_{n-4})\hspace{0.12em}=\hspace{0.12em}2, c(un−4un−3)=c(un−1un)=c(vn−1vn)=c(vn−3vn−2)=c(vn−2vn−1)=2c\left({u}_{n-4}{u}_{n-3})\hspace{0.12em}=\hspace{0.12em}c\left({u}_{n-1}{u}_{n})\hspace{0.12em}=\hspace{0.12em}c\left({v}_{n-1}{v}_{n})\hspace{0.12em}=\hspace{0.12em}c\left({v}_{n-3}{v}_{n-2})=c({v}_{n-2}{v}_{n-1})=2, c(un−3un−2)=c(un−2un−1)=c(vn−4vn−3)=c(u2v2)=1c\left({u}_{n-3}{u}_{n-2})=c\left({u}_{n-2}{u}_{n-1})=c\left({v}_{n-4}{v}_{n-3})=c\left({u}_{2}{v}_{2})=1, c(u1u1)=c(u1un)=c(v1vn)=3c\left({u}_{1}{u}_{1})=c\left({u}_{1}{u}_{n})=c\left({v}_{1}{v}_{n})=3; c(u4jv4j)=1c\left({u}_{4j}{v}_{4j})=1, c(u4j−1v4j−1)=2c\left({u}_{4j-1}{v}_{4j-1})=2, c(u4j+1v4j+1)=3c\left({u}_{4j+1}{v}_{4j+1})=3, where 1≤j≤t−11\le j\le t-1. Note that u1u2⋯un−2vn−2vn−3⋯v1u1{u}_{1}{u}_{2}\cdots {u}_{n-2}{v}_{n-2}{v}_{n-3}\cdots {v}_{1}{u}_{1}is a total proper cycle, any two distinct vertices of the cycle have two disjoint total proper paths. Now, we may assume that x=uix={u}_{i}for 1≤i≤n−21\le i\le n-2. If y=un−1y={u}_{n-1}, then xui+1⋯un−2vn−2vn−1yx{u}_{i+1}\cdots {u}_{n-2}{v}_{n-2}{v}_{n-1}yand xui−1⋯u1unyx{u}_{i-1}\cdots {u}_{1}{u}_{n}yare two total proper paths connecting xxand yy. If y=uny={u}_{n}, then xui+1⋯un−2vn−2vn−1⋯v1vnyx{u}_{i+1}\cdots {u}_{n-2}{v}_{n-2}{v}_{n-1}\cdots {v}_{1}{v}_{n}yand xui−1⋯u1v1vnyx{u}_{i-1}\cdots {u}_{1}{v}_{1}{v}_{n}yare two total proper paths connecting xxand yy. If y=vn−1y={v}_{n-1}, then xui−1⋯u3v3v2v1vnyx{u}_{i-1}\cdots {u}_{3}{v}_{3}{v}_{2}{v}_{1}{v}_{n}yand xui+1⋯un−2vn−2yx{u}_{i+1}\cdots {u}_{n-2}{v}_{n-2}yare two total proper paths connecting xxand yy. If y=vny={v}_{n}, then xui−1⋯u1unyx{u}_{i-1}\cdots {u}_{1}{u}_{n}yand xui+1⋯un−2vn−2vn−1yx{u}_{i+1}\cdots {u}_{n-2}{v}_{n-2}{v}_{n-1}yare two total proper paths connecting xxand yy. Assume that x=vix={v}_{i}for 1≤i≤n−21\le i\le n-2. If y=vn−1y={v}_{n-1}, then xvi+1⋯vn−2un−2un−1⋯u1unun−1yx{v}_{i+1}\cdots {v}_{n-2}{u}_{n-2}{u}_{n-1}\cdots {u}_{1}{u}_{n}{u}_{n-1}yand xvi−1⋯v1vnyx{v}_{i-1}\cdots {v}_{1}{v}_{n}yare two total proper paths connecting xxand yy. If y=vny={v}_{n}, then xvi+1⋯vn−2un−2un−1unyx{v}_{i+1}\cdots {v}_{n-2}{u}_{n-2}{u}_{n-1}{u}_{n}yand xvi−1⋯v1yx{v}_{i-1}\cdots {v}_{1}yare two total proper paths connecting xxand yy. Thus, CL2n{{\rm{CL}}}_{2n}is total proper 2-connected with the above coloring.To the contrary, suppose there exists a total proper 3-connected coloring ccof CL2n{{\rm{CL}}}_{2n}using three colors. Considering u1{u}_{1}and v2{v}_{2}, u1u2v2,u1v1v2{u}_{1}{u}_{2}{v}_{2},{u}_{1}{v}_{1}{v}_{2}, and u1unvnvn−1⋯v3v2{u}_{1}{u}_{n}{v}_{n}{v}_{n-1}\cdots {v}_{3}{v}_{2}must be three total proper paths connecting u1{u}_{1}and v2{v}_{2}. Then c(u1u2)≠c(u2v2)≠c(u2)c\left({u}_{1}{u}_{2})\ne c\left({u}_{2}{v}_{2})\ne c\left({u}_{2}). Considering u1{u}_{1}and u3{u}_{3}, u1u2u3,u1unun−1⋯u4u3{u}_{1}{u}_{2}{u}_{3},{u}_{1}{u}_{n}{u}_{n-1}\cdots {u}_{4}{u}_{3}, and u1v1v2v3u3{u}_{1}{v}_{1}{v}_{2}{v}_{3}{u}_{3}must be three total proper paths connecting u1{u}_{1}and u3{u}_{3}. Hence, c(u1u2)≠c(u2u3)≠c(u2)c\left({u}_{1}{u}_{2})\ne c\left({u}_{2}{u}_{3})\ne c\left({u}_{2}), and so c(u2v2)=c(u2u3)c\left({u}_{2}{v}_{2})=c\left({u}_{2}{u}_{3}). But then, there is no set of three disjoint total proper paths connecting u3{u}_{3}and v2{v}_{2}, a contradiction. Hence, tpc3(CL2n)≥4{{\rm{tpc}}}_{3}\left({{\rm{CL}}}_{2n})\ge 4. Now we only need to prove tpc3(CL2n)≤4{{\rm{tpc}}}_{3}\left({{\rm{CL}}}_{2n})\le 4.Case 1. n≡0(mod3)n\equiv 0\left({\rm{mod}}\hspace{0.33em}3). Let n=3tn=3t. Assign a total coloring ccto CL2n{{\rm{CL}}}_{2n}as follows: Let iibe an integer with 0≤i≤t−10\hspace{-0.1em}\le i\hspace{-0.1em}\le \hspace{-0.1em}t\hspace{-0.1em}-\hspace{-0.1em}1, c(u3i+1)=c(u3i+2u3i+3)=c(v3i+2)=c(v3i+3v3i+4)=1c\left({u}_{3i+1})\hspace{-0.1em}=\hspace{-0.1em}c\left({u}_{3i+2}{u}_{3i+3})=c\left({v}_{3i+2})\hspace{-0.1em}=\hspace{-0.1em}c\left({v}_{3i+3}{v}_{3i+4})\hspace{-0.1em}=\hspace{-0.1em}1, c(u3i+2)=c(u3i+3u3i+4)=c(v3i+3)=c(v3i+1v3i+2)=2c\left({u}_{3i+2})\hspace{-0.1em}=\hspace{-0.1em}c\left({u}_{3i+3}{u}_{3i+4})\hspace{-0.1em}=\hspace{-0.1em}c\left({v}_{3i+3})\hspace{-0.1em}=\hspace{-0.1em}c\left({v}_{3i+1}{v}_{3i+2})\hspace{-0.1em}=\hspace{-0.1em}2, and c(u3i+3)=c(u3i+1u3i+2)=c(v3i+1)=c(v3i+2v3i+3)=3c\left({u}_{3i+3})=c\left({u}_{3i+1}{u}_{3i+2})=c\left({v}_{3i+1})=c\left({v}_{3i+2}{v}_{3i+3})=3; c(ujvj)=4c\left({u}_{j}{v}_{j})=4for 1≤j≤n1\le j\le n. Let xxand yybe any two distinct vertices of CL2n{{\rm{CL}}}_{2n}. By symmetry, we may assume that x=u1x={u}_{1}. If y=uiy={u}_{i}for 2≤i≤n2\le i\le n, then xu2u3⋯ui−1y,xunun−1⋯ui+1yx{u}_{2}{u}_{3}\cdots {u}_{i-1}y,x{u}_{n}{u}_{n-1}\cdots {u}_{i+1}y, and xv1v2⋯viyx{v}_{1}{v}_{2}\cdots {v}_{i}yare three total proper paths connecting xxand yy. If y=v1y={v}_{1}, then xy,xu2v2yxy,x{u}_{2}{v}_{2}yand xunvnyx{u}_{n}{v}_{n}yare three total proper paths connecting xxand yy. If y=viy={v}_{i}for 2≤i≤n2\le i\le n, then xu1u2⋯uiy,xv1v2⋯vi−1yx{u}_{1}{u}_{2}\cdots {u}_{i}y,x{v}_{1}{v}_{2}\cdots {v}_{i-1}y, and xunvnvn−1⋯uiyx{u}_{n}{v}_{n}{v}_{n-1}\cdots {u}_{i}yare three total proper paths connecting xxand yy. Hence, CL2n{{\rm{CL}}}_{2n}is total proper 3-connected with the above coloring.Case 2. n≡1(mod3)n\equiv 1\left({\rm{mod}}\hspace{0.33em}3). Let n=3t+1n=3t+1. Define a total coloring ccof CL2n{{\rm{CL}}}_{2n}as follows: c(u3i+1)=c(u3i+2u3i+3)=c(v3i+3)=c(v3i+1v3i+2)=1c\left({u}_{3i+1})=c\left({u}_{3i+2}{u}_{3i+3})=c\left({v}_{3i+3})=c\left({v}_{3i+1}{v}_{3i+2})=1, c(u3i+2)=c(u3i+3u3i+4)=c(v3i+2)=c(v3i+3v3i+4)=3c\left({u}_{3i+2})=c\left({u}_{3i+3}{u}_{3i+4})=c\left({v}_{3i+2})=c\left({v}_{3i+3}{v}_{3i+4})=3, and c(u3i+3)=c(u3i+1u3i+2)=c(v3i+1)=c(v3i+2v3i+3)=2c\left({u}_{3i+3})=c\left({u}_{3i+1}{u}_{3i+2})=c\left({v}_{3i+1})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{3i+2}{v}_{3i+3})\hspace{-0.08em}=\hspace{-0.08em}2, where 0≤i≤t−10\le i\le t-1; c(un)=c(v1vn)=1,c(unvn)=2c\left({u}_{n})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{1}{v}_{n})\hspace{-0.08em}=\hspace{-0.08em}1,c\left({u}_{n}{v}_{n})\hspace{-0.08em}=\hspace{-0.08em}2, c(vn)=c(u1un)=4c\left({v}_{n})\hspace{-0.08em}=\hspace{-0.08em}c\left({u}_{1}{u}_{n})\hspace{-0.08em}=\hspace{-0.08em}4, and c(ujvj)=4c\left({u}_{j}{v}_{j})=4for 2≤j≤n−12\le j\le n-1. Let xxand yybe any two distinct vertices of CL2n{{\rm{CL}}}_{2n}. By symmetry, we may assume that x=u1x={u}_{1}. If y=uiy={u}_{i}for 2≤i≤n2\le i\le n, then xu2u3⋯ui−1y,xunun−1⋯ui+1yx{u}_{2}{u}_{3}\cdots {u}_{i-1}y,x{u}_{n}{u}_{n-1}\cdots {u}_{i+1}y, and xv1v2⋯viyx{v}_{1}{v}_{2}\cdots {v}_{i}yare three total proper paths connecting xxand yy. If y=v1y={v}_{1}, then xy,xu2v2yxy,x{u}_{2}{v}_{2}yand xunvnyx{u}_{n}{v}_{n}yare three total proper paths connecting xxand yy. If y=viy={v}_{i}for 2≤i≤n2\le i\le n, then xu1u2⋯uiy,xv1v2⋯vi−1yx{u}_{1}{u}_{2}\cdots {u}_{i}y,x{v}_{1}{v}_{2}\cdots {v}_{i-1}y, and xunvnvn−1⋯uiyx{u}_{n}{v}_{n}{v}_{n-1}\cdots {u}_{i}yare three total proper paths connecting xxand yy. Hence, CL2n{{\rm{CL}}}_{2n}is total proper 3-connected with the above coloring.Case 3. n≡2(mod3)n\equiv 2\left({\rm{mod}}\hspace{0.33em}3). Let n=3t+2n=3t+2. Define a total coloring ccof CL2n{{\rm{CL}}}_{2n}as follows: c(u3i+1)=c(u3i+2u3i+3)=c(v3i+4)=c(v3i+2v3i+3)=1c\left({u}_{3i+1})=c\left({u}_{3i+2}{u}_{3i+3})=c\left({v}_{3i+4})=c\left({v}_{3i+2}{v}_{3i+3})=1, c(u3i+2)=c(u3i+3u3i+4)=c(v3i+3)=c(v3i+1v3i+2)=3c\left({u}_{3i+2})=c\left({u}_{3i+3}{u}_{3i+4})=c\left({v}_{3i+3})=c\left({v}_{3i+1}{v}_{3i+2})=3, and c(u3i+3)=c(u3i+1u3i+2)=c(v3i+2)=c(v3i+3v3i+4)=2c\left({u}_{3i+3})=c\left({u}_{3i+1}{u}_{3i+2})=c\left({v}_{3i+2})=c\left({v}_{3i+3}{v}_{3i+4})=2, where 0≤i≤t−10\le i\le t-1; c(un)=c(vn−1vn)=3,c(vn)=c(un−1un)=c(u1v1)=2c\left({u}_{n})=c\left({v}_{n-1}{v}_{n})=3,c\left({v}_{n})=c\left({u}_{n-1}{u}_{n})=c\left({u}_{1}{v}_{1})=2, c(un−1)=c(v1vn)=c(unu1)=1,c(v1)=4c\left({u}_{n-1})=c\left({v}_{1}{v}_{n})=c\left({u}_{n}{u}_{1})=1,c\left({v}_{1})=4, and c(ujvj)=4c\left({u}_{j}{v}_{j})=4for 2≤j≤n2\le j\le n. Let xxand yybe any two distinct vertices of CL2n{{\rm{CL}}}_{2n}. By symmetry, we may assume that x=u1x={u}_{1}. If y=uiy={u}_{i}for 2≤i≤n2\le i\le n, then xu2u3⋯ui−1y,xunun−1⋯ui+1yx{u}_{2}{u}_{3}\cdots {u}_{i-1}y,x{u}_{n}{u}_{n-1}\cdots {u}_{i+1}y, and xv1v2⋯viyx{v}_{1}{v}_{2}\cdots {v}_{i}yare three total proper paths connecting xxand yy. If y=v1y={v}_{1}, then xy,xu2v2yxy,x{u}_{2}{v}_{2}y, and xunvnyx{u}_{n}{v}_{n}yare three total proper paths connecting xxand yy. If y=viy={v}_{i}for 2≤i≤n2\le i\le n, then xu1u2⋯uiy,xv1v2⋯vi−1yx{u}_{1}{u}_{2}\cdots {u}_{i}y,x{v}_{1}{v}_{2}\cdots {v}_{i-1}yand xunvnvn−1⋯uiyx{u}_{n}{v}_{n}{v}_{n-1}\cdots {u}_{i}yare three total proper paths connecting xxand yy. Hence, CL2n{{\rm{CL}}}_{2n}is total proper 3-connected with the above coloring.□Next, we shall determine the total proper kk-connection numbers of the Möbius ladders.Lemma 7.2Let nnbe an integer with n≥3n\ge 3. Then tpc(M2n)=tpc2(M2n)=3,tpc3(M2n)=4{\rm{tpc}}\left({M}_{2n})={{\rm{tpc}}}_{2}\left({M}_{2n})=3,{{\rm{tpc}}}_{3}\left({M}_{2n})=4.ProofSince M2n{M}_{2n}contains a Hamiltonian path and is not complete, we have tpc(M2n)=3{\rm{tpc}}\left({M}_{2n})=3. Since tpc2(M2n)≥tpc(M2n)=3{{\rm{tpc}}}_{2}\left({M}_{2n})\ge {\rm{tpc}}\left({M}_{2n})=3, we only need to prove tpc2(M2n)≤3{{\rm{tpc}}}_{2}\left({M}_{2n})\le 3.Case 1. n≡0(mod3)n\equiv 0\left({\rm{mod}}\hspace{0.33em}3). Define a total coloring ccof M2n{M}_{2n}as follows: Let iibe an integer with 1≤i≤n−21\le i\le n-2, c(ui)=c(vn−i+1)=1,c(ui+1)=c(vn−i)=3c\left({u}_{i})=c\left({v}_{n-i+1})=1,c\left({u}_{i+1})=c\left({v}_{n-i})=3, and c(ui+2)=c(vn−i−1)=2c\left({u}_{i+2})=c\left({v}_{n-i-1})=2; c(uiui+1),c(ui),c(ui+1)∈{1,2,3}c\left({u}_{i}{u}_{i+1}),c\left({u}_{i}),c\left({u}_{i+1})\in \left\{1,2,3\right\}with c(uiui+1)≠c(ui)≠c(ui+1)c\left({u}_{i}{u}_{i+1})\ne c\left({u}_{i})\ne c\left({u}_{i+1})for 1≤i≤n−11\le i\le n-1; c(vivi+1),c(vi),c(vi+1)∈{1,2,3}c\left({v}_{i}{v}_{i+1}),c\left({v}_{i}),c\left({v}_{i+1})\in \left\{1,2,3\right\}with c(vivi+1)≠c(vi)≠c(vi+1)c\left({v}_{i}{v}_{i+1})\ne c\left({v}_{i})\ne c\left({v}_{i+1})for 1≤i≤n−11\le i\le n-1; c(u1v1)=3,c(unvn)=2c\left({u}_{1}{v}_{1})=3,c\left({u}_{n}{v}_{n})=2, and c(ujvn−j+1)=3c\left({u}_{j}{v}_{n-j+1})=3for 1≤j≤n1\le j\le n. Let xxand yybe any two distinct vertices of M2n{M}_{2n}. By symmetry, we may assume that x=u1x={u}_{1}. If y=uiy={u}_{i}for 2≤i≤n2\le i\le n, then xu2u3⋯ui−1yx{u}_{2}{u}_{3}\cdots {u}_{i-1}yand xv1v2⋯vnunun−1⋯ui+1yx{v}_{1}{v}_{2}\cdots {v}_{n}{u}_{n}{u}_{n-1}\cdots {u}_{i+1}yare two total proper paths connecting xxand yy. If y=v1y={v}_{1}, then xyxyand xu2u3yx{u}_{2}{u}_{3}yare two total proper paths connecting xxand yy. If y=viy={v}_{i}for 2≤i≤n2\le i\le n, then xu1u2⋯unvnvn−1⋯vi+1yx{u}_{1}{u}_{2}\cdots {u}_{n}{v}_{n}{v}_{n-1}\cdots {v}_{i+1}yand xv1v2⋯vi−1yx{v}_{1}{v}_{2}\cdots {v}_{i-1}yare two total proper paths connecting xxand yy. Thus, M2n{M}_{2n}is total proper 2-connected with the above coloring.Case 2. n≡1(mod3)n\equiv 1\left({\rm{mod}}\hspace{0.33em}3). Let n=3t+1n=3t+1. Define a total coloring ccof M2n{M}_{2n}as follows: Let iibe an integer with 0≤i≤t−10\le i\le t-1, c(u3i+1)=c(u3i+2u3i+3)=c(v3i+3)=c(v3i+1v3i+2)=1c\left({u}_{3i+1})=c\left({u}_{3i+2}{u}_{3i+3})=c\left({v}_{3i+3})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{3i+1}{v}_{3i+2})\hspace{-0.08em}=\hspace{-0.08em}1, c(u3i+2)=c(u3i+3u3i+4)=c(v3i+1)=c(v3i+2v3i+3)=3c\left({u}_{3i+2})\hspace{-0.08em}=\hspace{-0.08em}c\left({u}_{3i+3}{u}_{3i+4})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{3i+1})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{3i+2}{v}_{3i+3})=3, and c(u3i+3)=c(u3i+1u3i+2)=c(v3i+2)=c(v3i+3v3i+4)=2c\left({u}_{3i+3})=c\left({u}_{3i+1}{u}_{3i+2})=c\left({v}_{3i+2})=c\left({v}_{3i+3}{v}_{3i+4})=2; if j=3ij=3ifor 1≤i≤t1\le i\le t, then c(ujvn−j+1)=3c\left({u}_{j}{v}_{n-j+1})=3; if j≠3ij\ne 3ifor 1≤i≤t1\le i\le t, then c(ujvn−j+1),c(uj),c(vn−j+1)∈{1,2,3}c\left({u}_{j}{v}_{n-j+1}),c\left({u}_{j}),c\left({v}_{n-j+1})\in \left\{1,2,3\right\}with c(ujvn−j+1)≠c(uj)≠c(vn−j+1)c\left({u}_{j}{v}_{n-j+1})\ne c\left({u}_{j})\ne c\left({v}_{n-j+1}); c(un)=1,c(unvn)=c(u1v1)=2,c(vn)=3c\left({u}_{n})=1,c\left({u}_{n}{v}_{n})=c\left({u}_{1}{v}_{1})=2,c\left({v}_{n})=3. Let xxand yybe any two distinct vertices of M2n{M}_{2n}. We may assume that x=uix={u}_{i}for 1≤i≤n1\le i\le n. If y=ujy={u}_{j}for i≤j≤ni\le j\le n, then xui+1ui+2⋯uj−1yx{u}_{i+1}{u}_{i+2}\cdots {u}_{j-1}yand xui−1ui−2⋯u1vnunun−1⋯uj+1yx{u}_{i-1}{u}_{i-2}\cdots {u}_{1}{v}_{n}{u}_{n}{u}_{n-1}\cdots {u}_{j+1}yare two total proper paths connecting xxand yy. If y=vjy={v}_{j}for 1≤j≤n1\le j\le n, then xui−1ui−2⋯u1vnvn−1⋯vj+1yx{u}_{i-1}{u}_{i-2}\cdots {u}_{1}{v}_{n}{v}_{n-1}\cdots {v}_{j+1}yand xui+1⋯un−j+1yx{u}_{i+1}\cdots {u}_{n-j+1}yare two total proper paths connecting xxand yy. Thus, M2n{M}_{2n}is total proper 2-connected with the above coloring.Case 3. n≡2(mod3)n\equiv 2\left({\rm{mod}}\hspace{0.33em}3). Let n=3t+2n=3t+2. Define a total coloring ccof M2n{M}_{2n}as follows: Let iibe an integer with 0≤i≤t−10\hspace{0.08em}\le \hspace{0.08em}i\le t\hspace{0.08em}-\hspace{0.08em}1, c(u3i+1)=c(u3i+2u3i+3)=c(v3i+3)=c(v3i+1v3i+2)=c(u3i+2vn−3i−1)=1c\left({u}_{3i+1})\hspace{0.08em}=\hspace{0.08em}c\left({u}_{3i+2}{u}_{3i+3})\hspace{0.08em}=\hspace{0.08em}c\left({v}_{3i+3})=c\left({v}_{3i+1}{v}_{3i+2})=c\left({u}_{3i+2}{v}_{n-3i-1})=1, c(u3i+2)=c(u3i+3u3i+4)=c(v3i+1)=c(v3i+2v3i+3)=c(u3i+3vn−3i−2)=3c\left({u}_{3i+2})=c\left({u}_{3i+3}{u}_{3i+4})=c\left({v}_{3i+1})=c\left({v}_{3i+2}{v}_{3i+3})\hspace{-0.08em}=\hspace{-0.08em}c\left({u}_{3i+3}{v}_{n-3i-2})\hspace{-0.08em}=\hspace{-0.08em}3, and c(u3i+3)=c(u3i+1u3i+2)=c(v3i+2)=c(v3i+3v3i+4)=c(u3i+4vn−3i−3)=2c\left({u}_{3i+3})=c\left({u}_{3i+1}{u}_{3i+2})=c\left({v}_{3i+2})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{3i+3}{v}_{3i+4})\hspace{-0.08em}=\hspace{-0.08em}c\left({u}_{3i+4}{v}_{n-3i-3})=2; c(un−1)=c(unv1)=c(vn−1vn)=1c\left({u}_{n-1})=c\left({u}_{n}{v}_{1})=c\left({v}_{n-1}{v}_{n})=1, c(un)=c(vn−1)=c(u1vn)=3c\left({u}_{n})=c\left({v}_{n-1})=c\left({u}_{1}{v}_{n})=3, c(vn)=c(un−1un)=2c\left({v}_{n})=c\left({u}_{n-1}{u}_{n})=2. Let xxand yybe any two distinct vertices of M2n{M}_{2n}. We may assume that x=uix={u}_{i}for 1≤i≤n1\le i\le n. If y=ujy={u}_{j}for i≤j≤ni\le j\le n, then xui+1ui+2⋯uj−1yx{u}_{i+1}{u}_{i+2}\cdots {u}_{j-1}yand xui−1ui−2⋯u1vnunun−1⋯uj+1yx{u}_{i-1}{u}_{i-2}\cdots {u}_{1}{v}_{n}{u}_{n}{u}_{n-1}\cdots {u}_{j+1}yare two total proper paths connecting xxand yy. If y=vjy={v}_{j}for 1≤j≤n1\le j\le n, then xui−1ui−2⋯u1vnvn−1⋯vj+1yx{u}_{i-1}{u}_{i-2}\cdots {u}_{1}{v}_{n}{v}_{n-1}\cdots {v}_{j+1}yand xui+1⋯un−j+1yx{u}_{i+1}\cdots {u}_{n-j+1}yare two total proper paths connecting xxand yy. Thus, M2n{M}_{2n}is total proper 2-connected with the above coloring.To the contrary, suppose there exists a total proper 3-connected coloring ccof M2n{M}_{2n}using three colors. By considering the pair {u2,vn}\left\{{u}_{2},{v}_{n}\right\}, u2u3⋯unvn,u2u1vn{u}_{2}{u}_{3}\cdots {u}_{n}{v}_{n},{u}_{2}{u}_{1}{v}_{n}, and u2vn−1vn{u}_{2}{v}_{n-1}{v}_{n}must be three total proper paths connecting u2{u}_{2}and vn{v}_{n}. Then c(u2vn−1)≠c(vn−1vn)≠c(vn−1)c\left({u}_{2}{v}_{n-1})\ne c\left({v}_{n-1}{v}_{n})\ne c\left({v}_{n-1}). By considering the pair {u2,vn−2}\left\{{u}_{2},{v}_{n-2}\right\}, u2u1vnvn−1vn−2,u2u3vn−2{u}_{2}{u}_{1}{v}_{n}{v}_{n-1}{v}_{n-2},{u}_{2}{u}_{3}{v}_{n-2}, and u2vn−1vn{u}_{2}{v}_{n-1}{v}_{n}must be three total proper paths connecting u2{u}_{2}and vn−2{v}_{n-2}. Thus, c(u2vn−1)≠c(vn−1vn−2)≠c(vn−1)c\left({u}_{2}{v}_{n-1})\ne c\left({v}_{n-1}{v}_{n-2})\ne c\left({v}_{n-1}), and hence c(vn−1vn−2)=c(vn−1vn)c\left({v}_{n-1}{v}_{n-2})=c\left({v}_{n-1}{v}_{n}). But then, there is no set of three disjoint total proper paths connecting vn−2{v}_{n-2}and vn{v}_{n}, a contradiction. Thus, tpc3(M2n)≥4{{\rm{tpc}}}_{3}\left({M}_{2n})\ge 4. Now we only need to prove tpc3(M2n)≤4{{\rm{tpc}}}_{3}\left({M}_{2n})\le 4.Case 1. n≡0(mod2)n\equiv 0\left({\rm{mod}}\hspace{0.33em}2). Let n=2tn=2t. Assign a total coloring ccof M2n{M}_{2n}as follows: c(u2i+1)=c(vn−2i)=c(u2i+2vn−2i−1)=1c\left({u}_{2i+1})=c\left({v}_{n-2i})=c\left({u}_{2i+2}{v}_{n-2i-1})\hspace{-0.1em}=\hspace{-0.1em}1, c(u2i+2)=c(un−2i−1)=c(u2i+1vn−2i)=3c\left({u}_{2i+2})=c\left({u}_{n-2i-1})=c\left({u}_{2i+1}{v}_{n-2i})=3, c(vn−2ivn−2i−1)=c(u2i+1u2i+2)=2c\left({v}_{n-2i}{v}_{n-2i-1})\hspace{-0.1em}=\hspace{-0.1em}c\left({u}_{2i+1}{u}_{2i+2})=2, where 0≤i≤t−10\le i\le t-1; c(u2i+2u2i+3)=c(vn−2i−1vn−2i−2)=4c\left({u}_{2i+2}{u}_{2i+3})=c\left({v}_{n-2i-1}{v}_{n-2i-2})=4for 0≤i≤t−20\le i\le t-2; c(unvn)=c(u1v1)=4c\left({u}_{n}{v}_{n})=c\left({u}_{1}{v}_{1})=4. Let xxand yybe any two distinct vertices of M2n{M}_{2n}. By symmetry, we may assume that x=u1x={u}_{1}. If y=uiy={u}_{i}for 2≤i≤n2\le i\le n, then xu2⋯ui−1y,xvnunun−1⋯ui+1yx{u}_{2}\cdots {u}_{i-1}y,x{v}_{n}{u}_{n}{u}_{n-1}\cdots {u}_{i+1}y, and xv1v2⋯vn−i+1uix{v}_{1}{v}_{2}\cdots {v}_{n-i+1}{u}_{i}are three total proper paths connecting xxand yy. If y=v1y={v}_{1}, then xy,xu2u3⋯unyxy,x{u}_{2}{u}_{3}\cdots {u}_{n}y, and xvnvn−1⋯v2yx{v}_{n}{v}_{n-1}\cdots {v}_{2}yare three total proper paths connecting xxand yy. If y=viy={v}_{i}for 2≤i≤n2\le i\le n, then xv1v2⋯vi−1y,xvnvn−1⋯vi+1yx{v}_{1}{v}_{2}\cdots {v}_{i-1}y,x{v}_{n}{v}_{n-1}\cdots {v}_{i+1}yand xu2u3⋯un−i+1yx{u}_{2}{u}_{3}\cdots {u}_{n-i+1}yare three total proper paths connecting xxand yy. Hence, M2n{M}_{2n}is total proper 3-connected with the above coloring.Case 2. n≡1(mod2)n\equiv 1\left({\rm{mod}}\hspace{0.33em}2). For n=3n=3, we assign a total coloring ccto M6{M}_{6}as follows: c(u1)=c(v3)=c(u2u3)=c(v1v2)=1c\left({u}_{1})=c\left({v}_{3})=c\left({u}_{2}{u}_{3})=c\left({v}_{1}{v}_{2})\hspace{-0.1em}=\hspace{-0.1em}1, c(u2)=c(v2)=c(u3v3)=c(u1v1)=2c\left({u}_{2})=c\left({v}_{2})=c\left({u}_{3}{v}_{3})=c\left({u}_{1}{v}_{1})=2, c(u3)=c(v1)=c(u1u2)=c(v2v3)=3c\left({u}_{3})=c\left({v}_{1})=c\left({u}_{1}{u}_{2})\hspace{-0.1em}=\hspace{-0.1em}c\left({v}_{2}{v}_{3})=3, c(uivn−i+1)=4c\left({u}_{i}{v}_{n-i+1})=4for 1≤i≤31\le i\le 3. We can verify that M6{M}_{6}is total proper 3-connected with the above total coloring, and so tpc3(M6)=4{{\rm{tpc}}}_{3}\left({M}_{6})=4.For n=5n=5, define a total coloring ccof M10{M}_{10}as follows: c(u1)=c(v5)=1c\left({u}_{1})=c\left({v}_{5})=1, c(u2)=c(v4)=2c\left({u}_{2})=c\left({v}_{4})=2, c(u3)=c(v3)=c(u5)=c(v1)=3c\left({u}_{3})=c\left({v}_{3})=c\left({u}_{5})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{1})\hspace{-0.08em}=\hspace{-0.08em}3, c(u4)=c(v2)=4c\left({u}_{4})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{2})\hspace{-0.08em}=\hspace{-0.08em}4, c(u2u3)=c(u4u5)=c(v1v2)=c(v3v4)=1c\left({u}_{2}{u}_{3})\hspace{-0.1em}=\hspace{-0.1em}c\left({u}_{4}{u}_{5})\hspace{-0.1em}=\hspace{-0.1em}c\left({v}_{1}{v}_{2})\hspace{-0.1em}=\hspace{-0.1em}c\left({v}_{3}{v}_{4})\hspace{-0.1em}=\hspace{-0.1em}1, c(u3u4)=c(u5v5)=c(u1v1)=c(v2v3)=2c\left({u}_{3}{u}_{4})\hspace{-0.1em}=\hspace{-0.1em}c\left({u}_{5}{v}_{5})\hspace{-0.1em}=\hspace{-0.1em}c\left({u}_{1}{v}_{1})\hspace{-0.1em}=\hspace{-0.1em}c\left({v}_{2}{v}_{3})\hspace{-0.1em}=2, c(u4v2)=c(u1u2)=c(v4v5)=3c\left({u}_{4}{v}_{2})=c\left({u}_{1}{u}_{2})=c\left({v}_{4}{v}_{5})=3, c(uivn−i+1)=4c\left({u}_{i}{v}_{n-i+1})=4for i=1,2,3,5i=1,2,3,5. We can check that M10{M}_{10}is total proper 3-connected with the above total coloring, and so tpc3(M10)=4{{\rm{tpc}}}_{3}\left({M}_{10})=4.Subcase 2.1. Let n≡1(mod4)n\equiv 1\left({\rm{mod}}\hspace{0.33em}4)for n≥7n\ge 7. Let n=2t+1n=2t+1. Assign a total coloring ccto M2n{M}_{2n}as follows: c(u4i+1)=c(vn−4i)=1c\left({u}_{4i+1})=c\left({v}_{n-4i})=1, c(u4i+2)=c(vn−4i−1)=2c\left({u}_{4i+2})=c\left({v}_{n-4i-1})=2, c(u4i+3)=c(vn−4i−2)=3c\left({u}_{4i+3})=c\left({v}_{n-4i-2})=3, and c(u4i+4)=c(vn−4i−3)=4c\left({u}_{4i+4})=c\left({v}_{n-4i-3})=4, where 0≤i≤t−220\le i\le \frac{t-2}{2}; c(u4iu4i+1)=c(vn−4i+1vn−4i)=c(u4i+2vn−4i−1)=3c\left({u}_{4i}{u}_{4i+1})=c\left({v}_{n-4i+1}{v}_{n-4i})=c\left({u}_{4i+2}{v}_{n-4i-1})\hspace{-0.08em}=\hspace{-0.08em}3, c(u4i+1u4i+2)=c(vn−4ivn−4i−1)=c(u4i+3vn−4i−2)=4c\left({u}_{4i+1}{u}_{4i+2})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{n-4i}{v}_{n-4i-1})\hspace{-0.08em}=\hspace{-0.08em}c\left({u}_{4i+3}{v}_{n-4i-2})\hspace{-0.08em}=\hspace{-0.08em}4, c(u4i+2u4i+3)=c(vn−4i−1vn−4i−2)=c(u4ivn−4i+1)=1c\left({u}_{4i+2}{u}_{4i+3})\hspace{0.1em}=\hspace{0.1em}c\left({v}_{n-4i-1}{v}_{n-4i-2})=c\left({u}_{4i}{v}_{n-4i+1})\hspace{0.1em}=\hspace{0.1em}1, and c(vn−4i−2vn−4i−3)=c(u4i+3u4i+4)=c(u4i+1vn−4i)=2c\left({v}_{n-4i-2}{v}_{n-4i-3})\hspace{0.1em}=\hspace{0.1em}c\left({u}_{4i+3}{u}_{4i+4})\hspace{0.1em}=\hspace{0.1em}c\left({u}_{4i+1}{v}_{n-4i})\hspace{0.1em}=\hspace{0.1em}2, where 1≤i≤t−221\hspace{-0.08em}\le \hspace{-0.08em}i\hspace{-0.08em}\le \hspace{-0.08em}\frac{t-2}{2}; c(un)=c(v1)=c(u1u2)=c(vnvn−1)=3c\left({u}_{n})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{1})\hspace{-0.08em}=\hspace{-0.08em}c\left({u}_{1}{u}_{2})\hspace{-0.08em}=\hspace{-0.1em}c\left({v}_{n}{v}_{n-1})=3, c(u2u3)=c(vn−1vn−2)=c(v1v2)=c(un−1un)=c(un−1v2)=1c\left({u}_{2}{u}_{3})=c\left({v}_{n-1}{v}_{n-2})=c\left({v}_{1}{v}_{2})=c\left({u}_{n-1}{u}_{n})=c\left({u}_{n-1}{v}_{2})=1, c(u3u4)=c(vn−2vn−3)=c(u1v1)=c(unvn)=2c\left({u}_{3}{u}_{4})=c\left({v}_{n-2}{v}_{n-3})=c\left({u}_{1}{v}_{1})=c\left({u}_{n}{v}_{n})=2, and c(u1vn)=c(u2vn−1)=c(u3vn−2)=4c\left({u}_{1}{v}_{n})=c\left({u}_{2}{v}_{n-1})=c\left({u}_{3}{v}_{n-2})=4. Let xxand yybe any two distinct vertices of M2n{M}_{2n}. By symmetry, we may assume that x=u1x={u}_{1}. If y=uiy={u}_{i}for 2≤i≤n2\le i\le n, then xu2u3⋯ui−1y,xvnunun−1⋯ui+1yx{u}_{2}{u}_{3}\cdots {u}_{i-1}y,x{v}_{n}{u}_{n}{u}_{n-1}\cdots {u}_{i+1}y, and xv1v2⋯vn−i+1uix{v}_{1}{v}_{2}\cdots {v}_{n-i+1}{u}_{i}are three total proper paths connecting xxand yy. If y=v1y={v}_{1}, then xy,xu2u3⋯unyxy,x{u}_{2}{u}_{3}\cdots {u}_{n}y, and xvnvn−1⋯v2yx{v}_{n}{v}_{n-1}\cdots {v}_{2}yare three total proper paths connecting xxand yy. If y=viy={v}_{i}for 2≤i≤n2\le i\le n, then xv1v2⋯vi−1y,xvnvn−1⋯vi+1yx{v}_{1}{v}_{2}\cdots {v}_{i-1}y,x{v}_{n}{v}_{n-1}\cdots {v}_{i+1}yand xu2u3⋯un−i+1yx{u}_{2}{u}_{3}\cdots {u}_{n-i+1}yare three total proper paths connecting xxand yy. Hence, M2n{M}_{2n}is total proper 3-connected with the above coloring.Subcase 2.2. Let n≡3(mod4)n\equiv 3\left({\rm{mod}}\hspace{0.33em}4)for n≥7n\ge 7. Let n=2t+1n=2t+1. Assign a total coloring ccto M2n{M}_{2n}as follows: Let iibe an integer with 0≤i≤t−320\le i\le \frac{t-3}{2}, c(u4i+1)=c(vn−4i)=1c\left({u}_{4i+1})=c\left({v}_{n-4i})=1, c(u4i+2)=c(vn−4i−1)=2c\left({u}_{4i+2})=c\left({v}_{n-4i-1})=2, c(u4i+3)=c(vn−4i−2)=3c\left({u}_{4i+3})=c\left({v}_{n-4i-2})=3, and c(u4i+4)=c(vn−4i−3)=4c\left({u}_{4i+4})=c\left({v}_{n-4i-3})=4. Let iibe an integer with 1≤i≤t−121\le i\le \frac{t-1}{2}, c(u4iu4i+1)=c(vn−4i−1vn−4i)=c(u4i+2vn−4i−1)=3c\left({u}_{4i}{u}_{4i+1})=c\left({v}_{n-4i-1}{v}_{n-4i})=c\left({u}_{4i+2}{v}_{n-4i-1})=3, c(u4i+1u4i+2)=c(vn−4ivn−4i−1)=c(u4i+3vn−4i−2)=4c\left({u}_{4i+1}{u}_{4i+2})\hspace{0.1em}=\hspace{0.1em}c\left({v}_{n-4i}{v}_{n-4i-1})\hspace{0.1em}=\hspace{0.1em}c\left({u}_{4i+3}{v}_{n-4i-2})\hspace{0.1em}=\hspace{0.1em}4, c(u4i+2u4i+3)=c(u4ivn−4i+1)=c(vn−4i−1vn−4i−2)=1c\left({u}_{4i+2}{u}_{4i+3})\hspace{0.1em}=\hspace{0.1em}c\left({u}_{4i}{v}_{n-4i+1})\hspace{0.12em}=\hspace{0.12em}c\left({v}_{n-4i-1}{v}_{n-4i-2})\hspace{0.12em}=\hspace{0.12em}1, and c(u4i+1vn−4i)=c(u4i−1u4i)=c(vn−4i+2vn−4i+1)=2c\left({u}_{4i+1}{v}_{n-4i})\hspace{-0.1em}=\hspace{-0.1em}c\left({u}_{4i-1}{u}_{4i})\hspace{-0.1em}=\hspace{-0.1em}c\left({v}_{n-4i+2}{v}_{n-4i+1})\hspace{-0.1em}=\hspace{-0.1em}2; c(un)=c(v1)=c(u1u2)=c(vnvn−1)=3c\left({u}_{n})\hspace{-0.1em}=\hspace{-0.1em}c\left({v}_{1})\hspace{-0.1em}=\hspace{-0.1em}c\left({u}_{1}{u}_{2})\hspace{-0.1em}=\hspace{-0.1em}c\left({v}_{n}{v}_{n-1})\hspace{-0.1em}=\hspace{-0.1em}3, c(un−1)=c(v2)=c(u1v1)=c(unvn)=2c\left({u}_{n-1})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{2})\hspace{-0.08em}=c\left({u}_{1}{v}_{1})\hspace{-0.08em}=c\left({u}_{n}{v}_{n})=2, and c(un−2)=c(v3)=c(u2u3)=c(vn−1vn−2)=1c\left({u}_{n-2})=c\left({v}_{3})=c\left({u}_{2}{u}_{3})=c\left({v}_{n-1}{v}_{n-2})=1. By the similar proof of the above subcase, we can verify M2n{M}_{2n}is total proper 3-connected with the above coloring.□8Proof of Theorem 1.7Note that K3□K2=CL6,K3,3=M6,Q3=CL8{K}_{3}\hspace{0.33em}\square \hspace{0.33em}{K}_{2}={{\rm{CL}}}_{6},{K}_{3,3}={M}_{6},{Q}_{3}={{\rm{CL}}}_{8}and M8{M}_{8}. By means of Theorem 1.6, we can obtain their total proper kk-connection numbers. Now, we only need to consider K4,F1,F2,F3{K}_{4},{F}_{1},{F}_{2},{F}_{3}.Lemma 8.1tpc(K4)=1,tpc2(K4)=3,tpc3(K4)=4{\rm{tpc}}\left({K}_{4})=1,{{\rm{tpc}}}_{2}\left({K}_{4})=3,{{\rm{tpc}}}_{3}\left({K}_{4})=4.ProofBy [26], we know that tpc(K4)=1{\rm{tpc}}\left({K}_{4})=1. Suppose tpc2(K4)=2{{\rm{tpc}}}_{2}\left({K}_{4})=2, then there is no set of two disjoint total proper paths connecting u1{u}_{1}and u2{u}_{2}, a contradiction. Thus, tpc2(K4)≥3{{\rm{tpc}}}_{2}\left({K}_{4})\ge 3. Let V(K4)={u1,u2,u3,u4}V\left({K}_{4})=\left\{{u}_{1},{u}_{2},{u}_{3},{u}_{4}\right\}, we assign a total coloring ccto K4{K}_{4}as follows: c(u1)=c(u2u3)=c(u3u4)=c(u2u4)=1c\left({u}_{1})=c\left({u}_{2}{u}_{3})=c\left({u}_{3}{u}_{4})=c\left({u}_{2}{u}_{4})=1, c(u2)=c(u4)=c(u1u3)=2c\left({u}_{2})=c\left({u}_{4})=c\left({u}_{1}{u}_{3})=2, c(u3)=c(u1u2)=c(u1u4)=3c\left({u}_{3})=c\left({u}_{1}{u}_{2})=c\left({u}_{1}{u}_{4})=3. We can verify that the K4{K}_{4}is total proper 2-connected, so tpc2(K4)=3{{\rm{tpc}}}_{2}\left({K}_{4})=3.Now, we suppose there exists a total proper 3-connected coloring ccof K4{K}_{4}using three colors. Considering u1{u}_{1}and u2{u}_{2}, u1u2{u}_{1}{u}_{2}, u1u4u2{u}_{1}{u}_{4}{u}_{2}, and u1u3u2{u}_{1}{u}_{3}{u}_{2}must be the three total proper paths connecting u1{u}_{1}and u2{u}_{2}. Then c(u1u4)≠c(u2u4)≠c(u4)c\left({u}_{1}{u}_{4})\ne c\left({u}_{2}{u}_{4})\ne c\left({u}_{4}). Considering u1{u}_{1}and u3{u}_{3}, u1u3{u}_{1}{u}_{3}, u1u2u3{u}_{1}{u}_{2}{u}_{3}, and u1u4u3{u}_{1}{u}_{4}{u}_{3}must be the three total proper paths connecting u1{u}_{1}and u3{u}_{3}. Thus, c(u1u4)≠c(u3u4)≠c(u4)c\left({u}_{1}{u}_{4})\ne c\left({u}_{3}{u}_{4})\ne c\left({u}_{4}), and so c(u2u4)=c(u4u3)c\left({u}_{2}{u}_{4})=c\left({u}_{4}{u}_{3}). But then, there is no set of three disjoint total proper paths connecting u2{u}_{2}and u3{u}_{3}, a contradiction. Hence, tpc3(K4)≥4{{\rm{tpc}}}_{3}\left({K}_{4})\ge 4. Define a total coloring ccof K4{K}_{4}as follows: c(u1)=c(u3)=1c\left({u}_{1})=c\left({u}_{3})=1, c(u2)=c(u4)=2c\left({u}_{2})=c\left({u}_{4})=2, c(u1u2)=c(u3u4)=4c\left({u}_{1}{u}_{2})=c\left({u}_{3}{u}_{4})=4, c(u1u4)=c(u2u3)=3c\left({u}_{1}{u}_{4})=c\left({u}_{2}{u}_{3})=3, c(u1u3)=2c\left({u}_{1}{u}_{3})=2, c(u2u4)=1c\left({u}_{2}{u}_{4})=1. We can easily check that K4{K}_{4}is total proper 3-connected, and so tpc3(K4)=4{{\rm{tpc}}}_{3}\left({K}_{4})=4.□Lemma 8.2tpc(F1)=tpc2(F1)=3,tpc3(F1)=4{\rm{tpc}}\left({F}_{1})={{\rm{tpc}}}_{2}\left({F}_{1})=3,{{\rm{tpc}}}_{3}\left({F}_{1})=4.Figure 4The total proper k-connected coloring of F1, F2 and F3.ProofSince F1{F}_{1}has a Hamiltonian path that is not complete, we know that tpc(F1)=3{\rm{tpc}}\left({F}_{1})=3. It is easy to verify that F1{F}_{1}is total proper 2-connected depicted in Figure 4(a), and so tpc2(F1)=3{{\rm{tpc}}}_{2}\left({F}_{1})=3. Now, we suppose there exists a total proper 3-connected coloring ccof F1{F}_{1}using three colors. By considering the pair {u2,u8}\left\{{u}_{2},{u}_{8}\right\}, u2u8,u2u1u8{u}_{2}{u}_{8},{u}_{2}{u}_{1}{u}_{8}, and u2u3⋯u8{u}_{2}{u}_{3}\cdots {u}_{8}must be the three total proper paths connecting u2{u}_{2}and u8{u}_{8}. Then c(u1u2)≠c(u1u8)≠c(u1)c\left({u}_{1}{u}_{2})\ne c\left({u}_{1}{u}_{8})\ne c\left({u}_{1}). By considering the pair {u5,u8}\left\{{u}_{5},{u}_{8}\right\}, u5u6u7u8,u5u1u8{u}_{5}{u}_{6}{u}_{7}{u}_{8},{u}_{5}{u}_{1}{u}_{8}, and u5u4u3u2u8{u}_{5}{u}_{4}{u}_{3}{u}_{2}{u}_{8}must be the three total proper paths connecting u5{u}_{5}and u8{u}_{8}. Then c(u1u5)≠c(u1u8)≠c(u1)c\left({u}_{1}{u}_{5})\ne c\left({u}_{1}{u}_{8})\ne c\left({u}_{1}), and hence c(u1u2)=c(u1u5)c\left({u}_{1}{u}_{2})=c\left({u}_{1}{u}_{5}). But then, there is no set of three disjoint total proper paths connecting u2{u}_{2}and u5{u}_{5}, a contradiction. Hence, tpc3(F1)≥4{{\rm{tpc}}}_{3}\left({F}_{1})\ge 4. By Figure 4(b), we know that F1{F}_{1}is total proper 3-connected, and so tpc3(F1)=4{{\rm{tpc}}}_{3}\left({F}_{1})=4.□Lemma 8.3tpc(F2)=tpc2(F2)=3{\rm{tpc}}\left({F}_{2})={{\rm{tpc}}}_{2}\left({F}_{2})=3.ProofSince F1{F}_{1}has a Hamiltonian path that is not complete, we know that tpc(F2)=3{\rm{tpc}}\left({F}_{2})=3. We can check that the coloring shown in Figure 4(c) is total proper 2-connected. Thus, tpc2(F2)=3{{\rm{tpc}}}_{2}\left({F}_{2})=3.□Lemma 8.4tpc(F3)=tpc2(F3)=3,tpc3(F3)=4{\rm{tpc}}\left({F}_{3})={{\rm{tpc}}}_{2}\left({F}_{3})=3,{{\rm{tpc}}}_{3}\left({F}_{3})=4.ProofSince F3{F}_{3}has a Hamiltonian path that is not complete, we know that tpc(F3)=3{\rm{tpc}}\left({F}_{3})=3. It is easy to check that the coloring shown in Figure 4(d) is total proper 2-connected using three colors. Thus, tpc2(F3)=3{{\rm{tpc}}}_{2}\left({F}_{3})=3. Now, we suppose there exists a total proper 3-connected coloring ccof F3{F}_{3}using three colors. Considering u2{u}_{2}and u8{u}_{8}, u2u8,u2u1u8{u}_{2}{u}_{8},{u}_{2}{u}_{1}{u}_{8}and u2u3⋯u8{u}_{2}{u}_{3}\cdots {u}_{8}must be the three total proper paths connecting u2{u}_{2}and u8{u}_{8}. Then c(u1u2)≠c(u1u8)≠c(u1)c\left({u}_{1}{u}_{2})\ne c\left({u}_{1}{u}_{8})\ne c\left({u}_{1}). Considering u5{u}_{5}and u8{u}_{8}, u5u6u7u8,u5u1u8{u}_{5}{u}_{6}{u}_{7}{u}_{8},{u}_{5}{u}_{1}{u}_{8}, and u5u4u3u2u8{u}_{5}{u}_{4}{u}_{3}{u}_{2}{u}_{8}must be the three total proper paths connecting u5{u}_{5}and u8{u}_{8}. Thus, c(u1u5)≠c(u1u8)≠c(u1)c\left({u}_{1}{u}_{5})\ne c\left({u}_{1}{u}_{8})\ne c\left({u}_{1}), and so c(u1u2)=c(u1u5)c\left({u}_{1}{u}_{2})=c\left({u}_{1}{u}_{5}). But then, there is no set of three disjoint total proper paths connecting u2{u}_{2}and u5{u}_{5}, a contradiction. Hence, tpc3(F3)≥4{{\rm{tpc}}}_{3}\left({F}_{3})\ge 4. By Figure 4(e), we know that F3{F}_{3}is total proper 3-connected, and so tpc3(F3)=4{{\rm{tpc}}}_{3}\left({F}_{3})=4.□ http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Open Mathematics de Gruyter

Some results on the total proper k-connection number

Open Mathematics , Volume 20 (1): 15 – Jan 1, 2022

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de Gruyter
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© 2022 Yingbin Ma and Hui Zhang, published by De Gruyter
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2391-5455
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10.1515/math-2022-0025
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Abstract

1IntroductionIn this paper, all graphs under our consideration are simple, finite and undirected. We follow the notation and terminology of [1]. For a graph GG, we denote by V(G),E(G)V\left(G),E\left(G)and diam(G){\rm{diam}}\left(G)the vertex set, edge set and diameter of GG, respectively. The distance between two vertices uuand vvin a connected graph GG, denoted by dist(u,v){\rm{dist}}\left(u,v), is the length of a shortest path between them in GG. The eccentricity of a vertex vvin GGis defined as eccG(v)=maxx∈V(G)dist(v,x){{\rm{ecc}}}_{G}\left(v)={\max }_{x\in V\left(G)}{\rm{dist}}\left(v,x). For convenience, a set of internally pairwise vertex disjoint paths will be called disjoint.In recent years, colored notions of connectivity in graphs becomes a new and active subject in graph theory. Stating from rainbow connection [2], rainbow vertex connection [3] and total rainbow connection [4,5] appeared later. Many researchers are working in this field, and a lot of papers have been published in journals, see [6,7,8, 9,10,11, 12,13,14, 15,16] for details. The reader can also see [17] for a survey, [18] for a dynamic survey and [19] for a new monograph on this topic.In 2012, Borozan et al. [20] introduced the concept of proper kk-connection number. A path in an edge-colored graph is a proper path if any two adjacent edges on the path differ in color. An edge-colored graph is proper k-connected if any two distinct vertices of the graph are connected by kkdisjoint proper paths. The proper k-connection number of a kk-connected graph GG, denoted by pck(G)p{c}_{k}\left(G), is defined as the smallest number of colors that are needed in order to make GGproper kk-connected. For more results, the reader can see [21,22, 23,24] for details.As a natural generalization, Jiang et al. [25] presented the concept of proper vertex kk-connection number. A path in a vertex-colored graph is a vertex proper path if any two internal adjacent vertices of the path differ in color. A vertex-colored graph is proper vertex k-connected if any two distinct vertices of the graph are connected by kkdisjoint vertex proper paths. For a kk-connected graph GG, the proper vertex k-connection number of GG, denoted by pvck(G)pv{c}_{k}\left(G), is defined as the smallest number of colors required to make GGproper vertex kk-connected.Motivated by the concept of total chromatic number of graph, now for proper connection and proper vertex connection, the concept of total proper connection was introduced by Jiang et al. [26]. A total coloring of a graph GGis a mapping from the set V(G)∪E(G)V\left(G)\cup E\left(G)to some finite set of colors. A path in a total-colored graph is a total proper path if the coloring of the edges and internal vertices is proper, that is, any two adjacent or incident elements of edges and internal vertices on the path differ in color. A total-colored graph is total proper k-connected if any two distinct vertices of the graph are connected by kkdisjoint total proper paths. For a connected graph GG, the total proper k-connection number of a kk-connected graph GG, denoted by tpck(G){{\rm{tpc}}}_{k}\left(G), is defined as the smallest number of colors that are needed in order to make GGtotal proper kk-connected. For convenience, we write tpc(G){\rm{tpc}}\left(G)for tpc1(G){{\rm{tpc}}}_{1}\left(G). Obviously, tpc(G)≤tpc2(G)≤tpc3(G){\rm{tpc}}\left(G)\le {{\rm{tpc}}}_{2}\left(G)\le {{\rm{tpc}}}_{3}\left(G). By [26], if GGis complete, then tpc(G)=1{\rm{tpc}}\left(G)=1; if GGhas a Hamiltonian path that is not complete, then tpc(G)=3{\rm{tpc}}\left(G)=3. Note that if GGis a nontrivial connected graph and HHis a connected spanning subgraph of GG, then tpc(G)≤tpc(H){\rm{tpc}}\left(G)\le {\rm{tpc}}\left(H).In this paper, we investigate the total proper connection number of a graph GGunder some constraints on its complement graph G¯\overline{G}.Theorem 1.1Let GGbe a connected graph of order n≥3n\ge 3, if diam(G¯){\rm{diam}}\left(\overline{G})does not belong to the following two cases: (i) diam(G¯)=2,3{\rm{diam}}\left(\overline{G})=2,3, (ii) G¯\overline{G}contain exactly two components and one of them is trivial, then tpc(G)≤4{\rm{tpc}}\left(G)\le 4.For the remaining cases, tpc(G){\rm{tpc}}\left(G)can be very large as discussed in Section 2. Then we add a constraint, i.e., we let G¯\overline{G}be triangle-free. Hence, GGis claw-free, and we can derive our next main result:Theorem 1.2For a connected graph GG, if G¯\overline{G}is triangle-free, then tpc(G)=3{\rm{tpc}}\left(G)=3.Recall that a clique of a graph is a set of mutually adjacent vertices, and that a maximum clique is a clique of the largest possible size in a given graph. The clique number ω(G)\omega \left(G)of a graph GGis the number of vertices in a maximum clique in GG. Let GGbe a connected graph, and let XXbe a maximum clique of GG. We say that NX(u){N}_{X}\left(u)is the set of neighbors of uuin G[X]G\left[X]and dX(u)=∣NX(u)∣{d}_{X}\left(u)=| {N}_{X}\left(u)| . Let F=G[V(G)\X]F=G\left[V\left(G)\backslash X]. Kemnitz and Schiermeyer [9] considered graphs with rc(G)=2rc\left(G)=2and large clique number. In this paper, we characterize graphs with small total proper connection number with respect to their large clique number. If ω(G)=n\omega \left(G)=n, then GGis a complete graph, which implies tpc(G)=1{\rm{tpc}}\left(G)=1. If GGis connected and ω(G)=n−1\omega \left(G)=n-1, then GGhas a Hamiltonian path, and so tpc(G)=3{\rm{tpc}}\left(G)=3. For the cases ω(G)=n−2,n−3\omega \left(G)=n-2,n-3, we obtain the following three main results.Figure 1All connected cubic graphs of order 8.Theorem 1.3Let GGbe a connected graph of order nn. If ω(G)=n−2\omega \left(G)=n-2and XXis a maximum clique of GGwith V(G)\X={u1,u2}V\left(G)\backslash X\right=\left\{{u}_{1},{u}_{2}\right\}, then tpc(G)=3{\rm{tpc}}\left(G)=3.Theorem 1.4Let GGbe a connected graph of order nn, diam(G)=2{\rm{diam}}\left(G)=2. If ω(G)=n−3\omega \left(G)=n-3and XXis a maximum clique of GGwith V(G)\X={u1,u2,u3}V\left(G)\backslash X\right=\left\{{u}_{1},{u}_{2},{u}_{3}\right\}, then tpc(G)=3{\rm{tpc}}\left(G)=3or tpc(G)=4{\rm{tpc}}\left(G)=4for the following case F≅3K3,∣NX(u1)∩NX(u2)∩NX(u3)∣=1F\cong 3{K}_{3},| {N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{2})\cap {N}_{X}\left({u}_{3})| =1and dX(u1)=dX(u2)=dX(u3)=1{d}_{X}\left({u}_{1})={d}_{X}\left({u}_{2})={d}_{X}\left({u}_{3})=1.Theorem 1.5Let GGbe a connected graph of order nn, diam(G)≥3{\rm{diam}}\left(G)\ge 3. If ω(G)=n−3\omega \left(G)=n-3and XXis a maximum clique of GGwith V(G)\X={u1,u2,u3}V\left(G)\backslash X\right=\left\{{u}_{1},{u}_{2},{u}_{3}\right\}, then tpc(G)=3{\rm{tpc}}\left(G)=3, or tpc(G)=4{\rm{tpc}}\left(G)=4and one of the following holds. (i)F≅3K3,NX(u)∩NX(v)≠∅F\cong 3{K}_{3},{N}_{X}\left(u)\cap {N}_{X}\left(v)\ne \varnothing and dX(u)=dX(v)=1{d}_{X}\left(u)={d}_{X}\left(v)=1, where uuand vvare any two distinct vertices in V(G)\XV\left(G)\backslash X.(ii)F≅3K3,dX(u1)=dX(u2)=dX(u3)=1F\cong 3{K}_{3},{d}_{X}\left({u}_{1})={d}_{X}\left({u}_{2})={d}_{X}\left({u}_{3})=1and for any two vertices in V(G)\XV\left(G)\backslash X, there is no common neighbor in G[X]G\left[X].For an integer n≥3n\ge 3, the circular ladder CL2n{{\rm{CL}}}_{2n}of order 2n2nis a cubic graph constructed by taking two copies of the cycle Cn{C}_{n}on disjoint vertex sets (u1,u2,…,un)\left({u}_{1},{u}_{2},\ldots ,{u}_{n})and (v1,v2,…,vn)\left({v}_{1},{v}_{2},\ldots ,{v}_{n}), then joining the corresponding vertices uivi{u}_{i}{v}_{i}for 1≤i≤n1\le i\le n. The Möbius ladder M2n{M}_{2n}of order 2n2nis obtained from the ladder by deleting the edges u1un{u}_{1}{u}_{n}and v1vn{v}_{1}{v}_{n}, and then inserting two edges u1vn{u}_{1}{v}_{n}and unv1{u}_{n}{v}_{1}. Subscripts are considered modulo nn, and we can derive our next main result:Theorem 1.6Let nnbe an integer with n≥3n\ge 3. Then(i)tpc(CL2n)=tpc2(CL2n)=3{\rm{tpc}}\left({{\rm{CL}}}_{2n})={{\rm{tpc}}}_{2}\left({{\rm{CL}}}_{2n})=3, tpc3(CL2n)=4{{\rm{tpc}}}_{3}\left({{\rm{CL}}}_{2n})=4.(ii)tpc(M2n)=tpc2(M2n)=3{\rm{tpc}}\left({M}_{2n})={{\rm{tpc}}}_{2}\left({M}_{2n})=3, tpc3(M2n)=4{{\rm{tpc}}}_{3}\left({M}_{2n})=4.In [7], Fujie-Okamoto et al. investigated the rainbow kk-connection numbers of all small cubic graphs of order 8 or less. In this paper, we determine the total proper kk-connection numbers of all small cubic graphs of order 8 or less. We can easily verify that all such cubic graphs have orders 4, 6, or 8, and those with orders 4 or 6 are K4,K3,3{K}_{4},{K}_{3,3}, and K3□K2{K}_{3}\hspace{0.33em}\square \hspace{0.33em}{K}_{2}(where □\square denotes Cartesian product). In [27], it was shown that all connected cubic graphs of order 8 are Q3,M8{Q}_{3},{M}_{8}, F1,F2{F}_{1},{F}_{2}, and F3{F}_{3}, and these graphs are depicted in Figure 1. Our last main result is stated as follows:Theorem 1.7(i)tpc(K4)=1{\rm{tpc}}\left({K}_{4})=1, tpc2(K4)=3{{\rm{tpc}}}_{2}\left({K}_{4})=3, tpc3(K4)=4{{\rm{tpc}}}_{3}\left({K}_{4})=4.(ii)tpc(K3,3)=tpc2(K3,3)=3{\rm{tpc}}\left({K}_{3,3})={{\rm{tpc}}}_{2}\left({K}_{3,3})=3, tpc3(K3,3)=4{{\rm{tpc}}}_{3}\left({K}_{3,3})=4.(iii)tpc(K3□K2)=tpc2(K3□K2)=3{\rm{tpc}}\left({K}_{3}\hspace{0.33em}\square \hspace{0.33em}{K}_{2})={{\rm{tpc}}}_{2}\left({K}_{3}\hspace{0.33em}\square \hspace{0.33em}{K}_{2})=3, tpc3(K3□K2)=4{{\rm{tpc}}}_{3}\left({K}_{3}\hspace{0.33em}\square \hspace{0.33em}{K}_{2})=4.(iv)tpc(Q3)=tpc2(Q3)=3{\rm{tpc}}\left({Q}_{3})={{\rm{tpc}}}_{2}\left({Q}_{3})=3, tpc3(Q3)=4{{\rm{tpc}}}_{3}\left({Q}_{3})=4.(v)tpc(M8)=tpc2(M8)=3{\rm{tpc}}\left({M}_{8})={{\rm{tpc}}}_{2}\left({M}_{8})=3, tpc3(M8)=4{{\rm{tpc}}}_{3}\left({M}_{8})=4.(vi)tpc(F1)=tpc2(F1)=3{\rm{tpc}}\left({F}_{1})={{\rm{tpc}}}_{2}\left({F}_{1})=3, tpc3(F1)=4{{\rm{tpc}}}_{3}\left({F}_{1})=4.(vii)tpc(F2)=tpc2(F2)=3{\rm{tpc}}\left({F}_{2})={{\rm{tpc}}}_{2}\left({F}_{2})=3.(viii)tpc(F3)=tpc2(F3)=3{\rm{tpc}}\left({F}_{3})={{\rm{tpc}}}_{2}\left({F}_{3})=3, tpc3(F3)=4{{\rm{tpc}}}_{3}\left({F}_{3})=4.2Proof of Theorem 1.1In order to prove Theorem 1.1, we need the following lemmas.Lemma 2.1[26] For 2≤m≤n2\le m\le n, we have tpc(Km,n)=3{\rm{tpc}}\left({K}_{m,n})=3.Lemma 2.2[26] If GGis a complete multipartite graph that is neither a complete graph nor a tree, then tpc(G)=3{\rm{tpc}}\left(G)=3.Let NG¯i(x)={v:distG¯(x,v)=i}{N}_{\overline{G}}^{i}\left(x)=\left\{v:{{\rm{dist}}}_{\overline{G}}\left(x,v)=i\right\}, where 0≤i≤30\le i\le 3, and NG¯4(x)={v:dist(x,v)≥4}{N}_{\overline{G}}^{4}\left(x)=\left\{v:{\rm{dist}}\left(x,v)\ge 4\right\}. In this paper, we use NG¯i{N}_{\overline{G}}^{i}instead of NG¯i(x){N}_{\overline{G}}^{i}\left(x)for convenience. Then NG¯0={x}{N}_{\overline{G}}^{0}=\left\{x\right\}and NG¯1=NG¯(x){N}_{\overline{G}}^{1}={N}_{\overline{G}}\left(x).Lemma 2.3For a connected graph GG, if G¯\overline{G}is connected and diam(G¯)≥4{\rm{diam}}\left(\overline{G})\ge 4, then tpc(G)≤4{\rm{tpc}}\left(G)\le 4.ProofChoose a vertex xxwith eccG¯(x)=diam(G¯){{\rm{ecc}}}_{\overline{G}}\left(x)={\rm{diam}}\left(\overline{G}). By the definition of NG¯i{N}_{\overline{G}}^{i}, we know uv∈E(G)uv\in E\left(G)for any u∈NG¯i,v∈NG¯ju\in {N}_{\overline{G}}^{i},v\in {N}_{\overline{G}}^{j}with ∣i−j∣≥2| i-j| \ge 2. Now we define a total coloring of GGas follows: assign color 1 to the edges xuxufor u∈NG¯2u\in {N}_{\overline{G}}^{2}, all edges between NG¯1{N}_{\overline{G}}^{1}and NG¯3{N}_{\overline{G}}^{3}, and all vertices and edges in NG¯4{N}_{\overline{G}}^{4}; assign color 2 to the edges xuxufor u∈NG¯3u\in {N}_{\overline{G}}^{3}, all edges between NG¯2{N}_{\overline{G}}^{2}and NG¯4{N}_{\overline{G}}^{4}, and all vertices and edges in NG¯1{N}_{\overline{G}}^{1}; assign color 3 to the edges xuxufor u∈NG¯4u\in {N}_{\overline{G}}^{4}, all edges between NG¯1{N}_{\overline{G}}^{1}and NG¯4{N}_{\overline{G}}^{4}, and all vertices and edges in NG¯2,NG¯3{N}_{\overline{G}}^{2},{N}_{\overline{G}}^{3}; assign color 4 to the vertex xx.We prove that there is a total proper path between any two vertices uuand vvof GG. It is trivial when uv∈E(G)uv\in E\left(G). Thus, we only need to consider the pairs u,v∈NG¯iu,v\in {N}_{\overline{G}}^{i}or u∈NG¯i,v∈NG¯i+1u\in {N}_{\overline{G}}^{i},v\in {N}_{\overline{G}}^{i+1}. Note that P=xx3x1x4x2P=x{x}_{3}{x}_{1}{x}_{4}{x}_{2}is a total proper path, where xi∈NG¯i{x}_{i}\in {N}_{\overline{G}}^{i}. By means of the path PP, we can find that uuand vvare connected by some total proper path for any u∈NG¯i,v∈NG¯i+1u\in {N}_{\overline{G}}^{i},v\in {N}_{\overline{G}}^{i+1}. If i=1i=1, then P=ux3xx2x4vP=u{x}_{3}x{x}_{2}{x}_{4}vis a total proper path, where xi∈NG¯i{x}_{i}\in {N}_{\overline{G}}^{i}. If i=2i=2, then P=uxx4vP=ux{x}_{4}vis a total proper path, where x4∈NG¯4{x}_{4}\in {N}_{\overline{G}}^{4}. If i=3i=3, then P=ux1x4x2xvP=u{x}_{1}{x}_{4}{x}_{2}xvis a total proper path, where xi∈NG¯i{x}_{i}\in {N}_{\overline{G}}^{i}. If i=4i=4, then P=uxx2vP=ux{x}_{2}vis a total proper path, where x2∈NG¯2{x}_{2}\in {N}_{\overline{G}}^{2}. Hence, tpc(G)≤4{\rm{tpc}}\left(G)\le 4.□Proof of Theorem 1.1Assume that G¯\overline{G}is connected. Since diam(G¯)≥4{\rm{diam}}\left(\overline{G})\ge 4, we have tpc(G)≤4{\rm{tpc}}\left(G)\le 4by Lemma 2.3. Assume that G¯\overline{G}is disconnected. By the assumption, we know that there exist either at least three connected components or exactly two nontrivial components. Let Gi¯\overline{{G}_{i}}be the components of G¯\overline{G}with ti=∣V(Gi¯)∣{t}_{i}=| V\left(\overline{{G}_{i}})| , where 1≤i≤h1\le i\le h. Then GGcontains a connected spanning subgraph Kt1,t2,…,th{K}_{{t}_{1},{t}_{2},\ldots ,{t}_{h}}, and we have tpc(G)≤tpc(Kt1,t2,…,th)=3{\rm{tpc}}\left(G)\le {\rm{tpc}}\left({K}_{{t}_{1},{t}_{2},\ldots ,{t}_{h}})=3by Lemma 2.2. Note that GGis not complete. Thus, tpc(G)≥3{\rm{tpc}}\left(G)\ge 3, and so tpc(G)=3{\rm{tpc}}\left(G)=3.□Next, we will give three examples to show that tpc(G){\rm{tpc}}\left(G)can be arbitrarily large if one of the following three conditions holds: diam(G¯)=2,diam(G¯)=3,G¯{\rm{diam}}\left(\overline{G})=2,{\rm{diam}}\left(\overline{G})=3,\overline{G}contains exactly two connected components and one of them is trivial.Figure 2The graph of Example 2.4.Example 2.4For the graph G¯\overline{G}shown in Figure 2, we choose a vertex xxwith eccG¯(x)=diam(G¯){{\rm{ecc}}}_{\overline{G}}\left(x)={\rm{diam}}\left(\overline{G}). Let NG¯1(x)={ui∣1≤i≤k}{N}_{\overline{G}}^{1}\left(x)=\left\{{u}_{i}| 1\le i\le k\right\}, NG¯2(x)={vj∣1≤j≤k}{N}_{\overline{G}}^{2}\left(x)=\left\{{v}_{j}| 1\le j\le k\right\}, and let E(G¯)={xui∣1≤i≤k}∪{ui1ui2∣1≤i1,i2≤k}∪{vj1vj2∣1≤j1,j2≤k}∪{uivj∣1≤i,j≤k}\{uivi∣1≤i≤k}E\left(\overline{G})=\left\{x{u}_{i}| 1\le i\le k\right\}\cup \left\{{u}_{{i}_{1}}{u}_{{i}_{2}}| 1\le {i}_{1},{i}_{2}\le k\right\}\cup \left\{{v}_{{j}_{1}}{v}_{{j}_{2}}| 1\le {j}_{1},{j}_{2}\le k\right\}\left\cup \left\{{u}_{i}{v}_{j}| 1\le i,j\le k\right\}\backslash \left\{{u}_{i}{v}_{i}| 1\le i\le k\right\}, where k≥3k\ge 3. Obviously, diam(G¯)=2{\rm{diam}}\left(\overline{G})=2and GGis a tree. Then tpc(G)=Δ(G)+1=k+1{\rm{tpc}}\left(G)=\Delta \left(G)+1=k+1by [26, Theorem 1]. Observe that tpc(G){\rm{tpc}}\left(G)will be arbitrarily large based on the increase of kk.Figure 3The graph of Example 2.5.Example 2.5For the graph G¯\overline{G}shown in Figure 3, we choose a vertex xxwith eccG¯(x)=diam(G¯){{\rm{ecc}}}_{\overline{G}}\left(x)={\rm{diam}}\left(\overline{G}). Let NG¯1(x)={ui∣1≤i≤k},NG¯2(x)={vj∣1≤j≤k}{N}_{\overline{G}}^{1}\left(x)=\left\{{u}_{i}| 1\le i\le k\right\},{N}_{\overline{G}}^{2}\left(x)=\left\{{v}_{j}| 1\le j\le k\right\}, and NG¯3(x)={ws∣1≤s≤k}{N}_{\overline{G}}^{3}\left(x)=\left\{{w}_{s}| 1\le s\le k\right\}, where k≥3k\ge 3. Furthermore, let E(G¯)={xui∣1≤i≤k}∪{uivj∣1≤i,j≤k}∪{vjws∣1≤j,s≤k}∪{vj1vj2∣1≤j1,j2≤k}E\left(\overline{G})=\left\{x{u}_{i}| 1\le i\le k\right\}\cup \left\{{u}_{i}{v}_{j}| 1\le i,j\le k\right\}\cup \left\{{v}_{j}{w}_{s}| 1\le j,s\le k\right\}\cup \left\{{v}_{{j}_{1}}{v}_{{j}_{2}}| 1\le {j}_{1},{j}_{2}\le k\right\}. Obviously, diam(G¯)=3{\rm{diam}}\left(\overline{G})=3and GGis a connected graph. Note that NG¯2(x){N}_{\overline{G}}^{2}\left(x)is a stable set in G¯\overline{G}, and each edge between xxand NG¯2(x){N}_{\overline{G}}^{2}\left(x)is a cut edge in GG. Therefore, tpc(G)≥k+1{\rm{tpc}}\left(G)\ge k+1by [26, Proposition 2], and so tpc(G){\rm{tpc}}\left(G)will be arbitrarily large based on the increase of kk.Example 2.6Let G¯\overline{G}contains exactly two components G1¯\overline{{G}_{1}}and G2¯\overline{{G}_{2}}, where G1¯\overline{{G}_{1}}is trivial and G2¯\overline{{G}_{2}}is a clique of G¯\overline{G}. Clearly, GGis a star, and tpc(G)=∣V(G2¯)∣+1{\rm{tpc}}\left(G)=| V\left(\overline{{G}_{2}})| +1. Thus, tpc(G){\rm{tpc}}\left(G)can be made arbitrarily large by increasing ∣V(G2¯)∣| V\left(\overline{{G}_{2}})| .3Proof of Theorem 1.2Lemma 3.1For a connected graph GG, if G¯\overline{G}is connected with diam(G¯)≥4{\rm{diam}}\left(\overline{G})\ge 4, and G¯\overline{G}is triangle-free, then tpc(G)=3{\rm{tpc}}\left(G)=3.ProofChoose a vertex xxwith eccG¯(x)=diam(G¯){{\rm{ecc}}}_{\overline{G}}\left(x)={\rm{diam}}\left(\overline{G}). Since G¯\overline{G}is triangle-free, we know that NG¯1{N}_{\overline{G}}^{1}is a clique in GG. Now we define a total coloring of GGas follows: assign color 1 to the edges xuxufor u∈NG¯2u\in {N}_{\overline{G}}^{2}, all edges between NG¯1{N}_{\overline{G}}^{1}and NG¯3{N}_{\overline{G}}^{3}, and all vertices and edges in NG¯4{N}_{\overline{G}}^{4}; assign color 2 to the edges between NG¯2{N}_{\overline{G}}^{2}and NG¯4{N}_{\overline{G}}^{4}, all vertices and edges in NG¯1{N}_{\overline{G}}^{1}, and the vertex xx; assign color 3 to the edges xuxufor u∈NG¯3,NG¯4u\in {N}_{\overline{G}}^{3},{N}_{\overline{G}}^{4}, all edges between NG¯1{N}_{\overline{G}}^{1}and NG¯4{N}_{\overline{G}}^{4}, and all vertices and edges in NG¯2,NG¯3{N}_{\overline{G}}^{2},{N}_{\overline{G}}^{3}.We prove that there is a total proper path between any two distinct vertices uuand vvin GG. Note that P=xx2x4x1x3P=x{x}_{2}{x}_{4}{x}_{1}{x}_{3}is a total proper path, where xi∈NG¯i{x}_{i}\in {N}_{\overline{G}}^{i}. By means of the path PP, we can find that uuand vvare connected by some total proper path for any u∈NG¯i,v∈NG¯i+1u\in {N}_{\overline{G}}^{i},v\in {N}_{\overline{G}}^{i+1}. Thus, we only need to consider the pairs u,v∈NG¯iu,v\in {N}_{\overline{G}}^{i}. For i=2i=2, P=uxx4vP=ux{x}_{4}vis a total proper path, where x4∈NG¯4{x}_{4}\in {N}_{\overline{G}}^{4}. For i=4i=4, P=uxx2vP=ux{x}_{2}vis a total proper path, where x2∈NG¯2{x}_{2}\in {N}_{\overline{G}}^{2}. For i=3i=3, P=ux1x4x2xvP=u{x}_{1}{x}_{4}{x}_{2}xvis a total proper path, where xi∈NG¯i{x}_{i}\in {N}_{\overline{G}}^{i}. Thus, GGis total proper connected with the above coloring, and so tpc(G)=3{\rm{tpc}}\left(G)=3.□Lemma 3.2Let GGbe a connected graph. If diam(G¯)=3{\rm{diam}}\left(\overline{G})=3and G¯\overline{G}is triangle-free, then tpc(G)=3{\rm{tpc}}\left(G)=3.ProofFor a vertex xxof G¯\overline{G}satisfying eccG¯(x)=diam(G¯)=3{{\rm{ecc}}}_{\overline{G}}\left(x)={\rm{diam}}\left(\overline{G})=3, let ni{n}_{i}represent the number of vertices with distance iifrom xx. If n1=n2=n3=1{n}_{1}={n}_{2}={n}_{3}=1, then G≅P4G\cong {P}_{4}, and so tpc(G)=3{\rm{tpc}}\left(G)=3.Case 1. Two of n1,n2,n3{n}_{1},{n}_{2},{n}_{3}are equal to 1. Without loss of generality, we may assume n1=n2=1{n}_{1}={n}_{2}=1. Since G¯\overline{G}is triangle-free, we have that NG¯3{N}_{\overline{G}}^{3}is a stable set in G¯\overline{G}, and so a clique in GG. We can find that GGhas a Hamiltonian path. Thus, tpc(G)=3{\rm{tpc}}\left(G)=3.Case 2. One of n1,n2,n3{n}_{1},{n}_{2},{n}_{3}is equal to 1. Suppose n2=1{n}_{2}=1. Since G¯\overline{G}is triangle-free, we know that NG¯1{N}_{\overline{G}}^{1}and NG¯3{N}_{\overline{G}}^{3}is a stable set in G¯\overline{G}, and so a clique in GG. Note that GGhas a Hamiltonian path, and so tpc(G)=3{\rm{tpc}}\left(G)=3.Subcase 2.1. n1=1{n}_{1}=1. Since G¯\overline{G}is triangle-free, we obtain that NG¯2{N}_{\overline{G}}^{2}is a clique in GG. Define a total coloring of GGas follows: assign color 3 to the vertex xx, all edges between NG¯2{N}_{\overline{G}}^{2}and NG¯3{N}_{\overline{G}}^{3}, and all edges between NG¯1{N}_{\overline{G}}^{1}and NG¯3{N}_{\overline{G}}^{3}; assign color 2 to the edges xuxufor u∈NG¯2u\in {N}_{\overline{G}}^{2}, and all vertices and edges in NG¯3{N}_{\overline{G}}^{3}; assign color 1 to the edges xuxufor u∈NG¯3u\in {N}_{\overline{G}}^{3}, and all vertices and edges in NG¯1,NG¯2{N}_{\overline{G}}^{1},{N}_{\overline{G}}^{2}. We prove that there is a total proper path between any two distinct vertices uuand vvin GG. Note that P=x1x3xx2P={x}_{1}{x}_{3}x{x}_{2}is a total proper path, where xi∈NG¯i{x}_{i}\in {N}_{\overline{G}}^{i}. By means of the path PP, we know that uuand vvare connected by some total proper path for any u∈NG¯i,v∈NG¯i+1u\in {N}_{\overline{G}}^{i},v\in {N}_{\overline{G}}^{i+1}. For any two vertices u,v∈NG¯3u,v\in {N}_{\overline{G}}^{3}, it is trivial if uv∈E(G)uv\in E\left(G). If uv∉E(G)uv\notin E\left(G), since u,v∈NG¯3u,v\in {N}_{\overline{G}}^{3}, there exist two vertices u′,v′∈NG¯3{u}^{^{\prime} },{v}^{^{\prime} }\in {N}_{\overline{G}}^{3}such that uu′,vv′∈E(G¯)u{u}^{^{\prime} },v{v}^{^{\prime} }\in E\left(\overline{G}). Since G¯\overline{G}is triangle-free, we can see that u′≠v′{u}^{^{\prime} }\ne {v}^{^{\prime} }and vu′,uv′∈E(G)v{u}^{^{\prime} },u{v}^{^{\prime} }\in E\left(G). Then P=uxu′vP=ux{u}^{^{\prime} }via a total proper path. Hence, GGis total proper connected with the above coloring, and so tpc(G)=3{\rm{tpc}}\left(G)=3.Subcase 2.2. n3=1{n}_{3}=1. Since G¯\overline{G}is triangle-free, we know that NG¯1{N}_{\overline{G}}^{1}is a stable set in G¯\overline{G}, and so a clique in GG. Define a total coloring of GGas follows: assign color 3 to the vertex xx, and all edges between NG¯1{N}_{\overline{G}}^{1}and NG¯3{N}_{\overline{G}}^{3}; assign color 2 to the edges xuxufor u∈NG¯2u\in {N}_{\overline{G}}^{2}, and all vertices and edges in NG¯3{N}_{\overline{G}}^{3}; assign color 1 to the edges xuxufor u∈NG¯3u\in {N}_{\overline{G}}^{3}, and all vertices and edges in NG¯1,NG¯2{N}_{\overline{G}}^{1},{N}_{\overline{G}}^{2}. We prove that there is a total proper path between any two distinct vertices uuand vvin GG. Note that P=x1x3xx2P={x}_{1}{x}_{3}x{x}_{2}is a total proper path, where xi∈NG¯i{x}_{i}\in {N}_{\overline{G}}^{i}. By means of the path PP, we obtain that uuand vvare connected by some total proper path for any u∈NG¯i,v∈NG¯i+1u\in {N}_{\overline{G}}^{i},v\in {N}_{\overline{G}}^{i+1}. Let u,vu,vbe any two distinct vertices of NG¯2{N}_{\overline{G}}^{2}, and NG¯3={y}{N}_{\overline{G}}^{3}=\{y\}. If yyis adjacent to any vertex of NG¯2{N}_{\overline{G}}^{2}in G¯\overline{G}, then NG¯2{N}_{\overline{G}}^{2}is a clique in GG, and so GGhas a Hamiltonian path. Otherwise, let Vy{V}_{y}denote the set of neighbors of yyin NG¯2{N}_{\overline{G}}^{2}in GG. We can check that P=uyxvP=uyxvis a total proper path, where u,v∈Vyu,v\in {V}_{y}. If ∣NG¯2\Vy∣=1\left| {N}_{\overline{G}}^{2}\backslash {V}_{y}\right| =1, then P=uyxvP=uyxvis a total proper path, where u∈Vy,v∈NG¯2\Vyu\in {V}_{y},v\left\in {N}_{\overline{G}}^{2}\backslash {V}_{y}. If ∣NG¯2\Vy∣≥2\left| {N}_{\overline{G}}^{2}\backslash {V}_{y}\right| \ge 2, then GGis claw-free since G¯\overline{G}is triangle-free, and G[x∪NG¯2\Vy]G\left[x\left\cup {N}_{\overline{G}}^{2}\backslash {V}_{y}]is a complete graph. Note that P=uyxvP=uyxvis a total proper path, where u∈Vy,v∈NG¯2u\in {V}_{y},v\in {N}_{\overline{G}}^{2}. Thus, GGis total proper connected with the above coloring, and so tpc(G)=3{\rm{tpc}}\left(G)=3.Case 3. n1,n2,n3≥2{n}_{1},{n}_{2},{n}_{3}\ge 2. Since G¯\overline{G}is triangle-free, we have that NG¯1{N}_{\overline{G}}^{1}is a stable set in G¯\overline{G}, and so a clique in GG. If any vertex in NG¯3{N}_{\overline{G}}^{3}is adjacent to all vertices of NG¯2{N}_{\overline{G}}^{2}in G¯\overline{G}, then both NG¯2{N}_{\overline{G}}^{2}and NG¯3{N}_{\overline{G}}^{3}are stable sets in G¯\overline{G}, and so cliques in GG. Thus, GGhas a Hamiltonian path, and so tpc(G)=3{\rm{tpc}}\left(G)=3.Otherwise, we choose a vertex u∈NG¯3u\in {N}_{\overline{G}}^{3}, let Vu{V}_{u}denote the set of neighbors of uuin NG¯2{N}_{\overline{G}}^{2}in GG, we have Vu≠∅,NG¯2{V}_{u}\ne \varnothing ,{N}_{\overline{G}}^{2}. Define a total coloring of GG: assign color 2 to the vertex xx, all vertices and edges in NG¯1{N}_{\overline{G}}^{1}, and all edges between NG¯2,NG¯3{N}_{\overline{G}}^{2},{N}_{\overline{G}}^{3}; assign color 3 to the vertex uu, all edges between NG¯1{N}_{\overline{G}}^{1}and NG¯3\{u}{N}_{\overline{G}}^{3}\backslash \left\{u\right\}, and all edges between xxand Vu{V}_{u}; assign color 2 to the remaining vertices and edges. Note that P=xvux1P=xvu{x}_{1}is a total proper path, where v∈Vu,x1∈NG¯1v\in {V}_{u},{x}_{1}\in {N}_{\overline{G}}^{1}. For any two vertices w,z∈NG¯3w,z\in {N}_{\overline{G}}^{3}, P=wx1uvxzP=w{x}_{1}uvxzis a total proper path, where x1∈NG¯1,v∈Vu{x}_{1}\in {N}_{\overline{G}}^{1},v\in {V}_{u}. For any two vertices w,z∈NG¯2w,z\in {N}_{\overline{G}}^{2}, P=wuxvP=wuxvis a total proper path, where u,v∈Vyu,v\in {V}_{y}. If ∣NG¯2\Vy∣=1\left| {N}_{\overline{G}}^{2}\backslash {V}_{y}\right| =1, then P=uxvP=uxvis a total proper path, where u∈Vy,v∈NG¯2\Vyu\in {V}_{y},v\left\in {N}_{\overline{G}}^{2}\backslash {V}_{y}. If ∣NG¯2\Vy∣≥2\left| {N}_{\overline{G}}^{2}\backslash {V}_{y}\right| \ge 2, since G¯\overline{G}is triangle-free, we know that GGis claw-free, and the subgraph G[x∪NG¯2\Vy]G\left[x\left\cup {N}_{\overline{G}}^{2}\backslash {V}_{y}]is a complete graph. Note that P=uxvP=uxvis a total proper path, where u∈Vy,v∈NG¯2\Vyu\in {V}_{y},v\left\in {N}_{\overline{G}}^{2}\backslash {V}_{y}. For any w∈NG¯2,z∈NG¯3w\in {N}_{\overline{G}}^{2},z\in {N}_{\overline{G}}^{3}, P=wxvux1zP=wxvu{x}_{1}zis a total proper path, where v∈Vy,x1∈NG¯1v\in {V}_{y},{x}_{1}\in {N}_{\overline{G}}^{1}. Similarly, there is a total proper path connecting any two vertices w∈NG¯2,z∈NG¯1w\in {N}_{\overline{G}}^{2},z\in {N}_{\overline{G}}^{1}. Hence, GGis total proper connected, and so tpc(G)=3{\rm{tpc}}\left(G)=3.□Lemma 3.3For a connected graph GG, if G¯\overline{G}is triangle-free and diam(G¯)=2{\rm{diam}}\left(\overline{G})=2, then tpc(G)=3{\rm{tpc}}\left(G)=3.ProofChoose a vertex xxwith eccG¯(x)=diam(G¯)=2{{\rm{ecc}}}_{\overline{G}}\left(x)={\rm{diam}}\left(\overline{G})=2. Since GGis connected, we have n1≥2,n2=1{n}_{1}\ge 2,\hspace{0.33em}{n}_{2}=1or n1,n2≥2{n}_{1},{n}_{2}\ge 2, and there exist two vertices u∈NG¯1,v∈NG¯2u\in {N}_{\overline{G}}^{1},v\in {N}_{\overline{G}}^{2}such that uv∈E(G)uv\in E\left(G). Assume n1≥2{n}_{1}\ge 2and n2=1{n}_{2}=1. Since G¯\overline{G}is triangle-free, we know that NG¯1{N}_{\overline{G}}^{1}is a stable set in G¯\overline{G}, and so a clique in GG. Note that GGhas a Hamiltonian path, and so tpc(G)=3{\rm{tpc}}\left(G)=3.Assume n1,n2≥2{n}_{1},{n}_{2}\ge 2. Observe that NG¯1{N}_{\overline{G}}^{1}is a stable set in G¯\overline{G}since G¯\overline{G}is triangle-free, and so a clique in GG. We show a total coloring of GGas follows: assign color 1 to the vertex xx, the edge uvuvand all vertices in NG¯1\u{N}_{\overline{G}}^{1}\backslash u; assign color 3 to the vertex vvand all edges in NG¯1{N}_{\overline{G}}^{1}; assign color 2 to the remaining vertices and edges. If there exist some vertices w∈NG¯2w\in {N}_{\overline{G}}^{2}with dG¯(w)=n−2{d}_{\overline{G}}\left(w)=n-2, then wwis adjacent to the remaining vertices except xxin G¯\overline{G}. Since diam(G¯)=2{\rm{diam}}\left(\overline{G})=2, there exists an edge w1w2∈E(G¯){w}_{1}{w}_{2}\in E\left(\overline{G})with w1∈NG¯1,w2∈NG¯2{w}_{1}\in {N}_{\overline{G}}^{1},{w}_{2}\in {N}_{\overline{G}}^{2}. Thus, w,w1,w2w,{w}_{1},{w}_{2}is a triangle in G¯\overline{G}, a contradiction. Hence, dG¯(w)<n−2{d}_{\overline{G}}\left(w)\lt n-2for all w∈NG¯2w\in {N}_{\overline{G}}^{2}, and so dG(w)≥2{d}_{G}\left(w)\ge 2. For any z∈NG¯1z\in {N}_{\overline{G}}^{1}, we know that P=xvuzP=xvuzis a total proper path. For any y∈NG¯2\{v}y\left\in {N}_{\overline{G}}^{2}\backslash \left\{v\right\}and z∈NG¯1z\in {N}_{\overline{G}}^{1}, if NG(y)∩NG¯1≠∅{N}_{G}(y)\cap {N}_{\overline{G}}^{1}\ne \varnothing , let w∈NG(y)∩NG¯1w\in {N}_{G}(y)\cap {N}_{\overline{G}}^{1}. Then ywzywzis a total proper path. Otherwise, let NG(y)∩NG¯1=∅{N}_{G}(y)\cap {N}_{\overline{G}}^{1}=\varnothing . We claim that yyis adjacent to all the other vertices of NG¯2{N}_{\overline{G}}^{2}in GG. In fact, for any vertex w∈NG¯2\{y}w\left\in {N}_{\overline{G}}^{2}\backslash \{y\}, there exists a vertex w′∈NG¯1w^{\prime} \in {N}_{\overline{G}}^{1}such that ww′∈E(G¯)w{w}^{^{\prime} }\in E\left(\overline{G}). Since yw′∈E(G¯)y{w}^{^{\prime} }\in E\left(\overline{G}), we know that yw∈E(G)yw\in E\left(G). Then yvuzyvuzis a total proper path. Next we consider w,z∈NG¯2w,z\in {N}_{\overline{G}}^{2}such that wz∉E(G)wz\notin E\left(G). Since G¯\overline{G}is triangle-free, we have that GGis claw-free, and at least one of wwand zzis adjacent to the vv, without loss of generality, assume that wv∈E(G)wv\in E\left(G). Since w,z∈NG¯2w,z\in {N}_{\overline{G}}^{2}, there exist two vertices w′,z′∈NG¯2{w}^{^{\prime} },{z}^{^{\prime} }\in {N}_{\overline{G}}^{2}such that ww′,zz′∈E(G¯)w{w}^{^{\prime} },z{z}^{^{\prime} }\in E\left(\overline{G}), and w′≠z′{w}^{^{\prime} }\ne {z}^{^{\prime} }. Then zw′,wz′∈E(G)z{w}^{^{\prime} },w{z}^{^{\prime} }\in E\left(G)and P=wvuw′zP=wvu{w}^{^{\prime} }zis a total proper path. Thus, GGis total proper connected with the above coloring. Hence, tpc(G)=3{\rm{tpc}}\left(G)=3.□Lemma 3.4Let GGbe a connected graph of order n≥3n\ge 3. If G¯\overline{G}is disconnected and triangle-free, then tpc(G)=3{\rm{tpc}}\left(G)=3.ProofSuppose G¯\overline{G}is triangle-free and contains two connected components one of which is trivial. Let G1¯\overline{{G}_{1}}and G2¯\overline{{G}_{2}}be the two components of G¯\overline{G}, where V(G1¯)={u}V\left(\overline{{G}_{1}})=\left\{u\right\}. Then uuis adjacent to any other vertex in GG. We will consider two cases according to the value of δ\delta , where δ\delta is the minimum degree of GG. If δ=1\delta =1, let d(v)=δd\left(v)=\delta . Since G¯\overline{G}is triangle-free, we know that GGis claw-free, and the subgraph G[V(G)\{v}]G\left[V\left(G)\backslash \left\{v\right\}]is a complete graph. Thus, GGhas a Hamiltonian path, and so tpc(G)=3{\rm{tpc}}\left(G)=3. If δ≥2\delta \ge 2, let d(v)=δ,D=V(G)\{u,v}d\left(v)=\delta ,D\left=V\left(G)\backslash \left\{u,v\right\}, and Vv{V}_{v}be the set of neighbors of vvin GG. Now we define a total coloring of GGas follows: assign color 1 to the vertex vvand all the edges between uuand Vv{V}_{v}; assign color 3 to the vertex uuand all the edges between vvand Vv{V}_{v}; assign color 2 to the remaining vertices and edges. Since GGis claw-free, we can find that the subgraph G[V(G)\{v}∪Vv]G\left[V\left(G)\backslash \left\{v\right\}\right\cup {V}_{v}]is a complete graph, and P=v1uvv2P={v}_{1}uv{v}_{2}is a total proper path, where v1,v2∈Vv{v}_{1},{v}_{2}\in {V}_{v}. For any w∈Vv,z∈D\Vvw\in {V}_{v},z\left\in D\backslash {V}_{v}, we obtain that P=wuzP=wuzis a total proper path. Thus, GGis total proper connected with the above coloring, and so tpc(G)=3{\rm{tpc}}\left(G)=3. Suppose G¯\overline{G}contains at least three connected components or exactly two nontrivial components. Then we have tpc(G)=3{\rm{tpc}}\left(G)=3by the similar proof of Theorem 1.1.□Proof of Theorem 1.2If G¯\overline{G}is connected, the result holds for the case diam(G¯)≥4{\rm{diam}}\left(\overline{G})\ge 4by Lemma 3.1, the case diam(G¯)=3{\rm{diam}}\left(\overline{G})=3by Lemma 3.2, and the case diam(G¯)=2{\rm{diam}}\left(\overline{G})=2by Lemma 3.3. If G¯\overline{G}is disconnected, the result holds by Lemma 3.4.□4Proof of Theorem 1.3Suppose F≅K2F\cong {K}_{2}. Note that GGhas a Hamiltonian path, and thus tpc(G)=3{\rm{tpc}}\left(G)=3. Next, we compute the total proper connection number of GGby proving the following claim.Claim 1. Let GGbe a graph obtained by adding two pendant vertices {u1,u2}\left\{{u}_{1},{u}_{2}\right\}to a vertex v1{v}_{1}of a complete graph Kt{K}_{t}. Then tpc(G)=3{\rm{tpc}}\left(G)=3.ProofSince GGis not a complete graph, we have tpc(G)≥3{\rm{tpc}}\left(G)\ge 3. Now we only need to prove tpc(G)≤3{\rm{tpc}}\left(G)\le 3by the following cases.Case 1. t≡0(mod3)t\equiv 0\left({\rm{mod}}\hspace{0.33em}3). Assign a total coloring ccto GGas follows: Let c(u1v1)=1,c(u2v1)=3c\left({u}_{1}{v}_{1})=1,c\left({u}_{2}{v}_{1})=3; c(v3i+1)=c(v3i+2v3i+3)=2c\left({v}_{3i+1})=c\left({v}_{3i+2}{v}_{3i+3})=2, c(v3i+2)=c(v3i+3v3i+4)=1c\left({v}_{3i+2})=c\left({v}_{3i+3}{v}_{3i+4})=1, and c(v3i+3)=c(v3i+1u3i+2)=3c\left({v}_{3i+3})=c\left({v}_{3i+1}{u}_{3i+2})=3, where 0≤i≤t3−10\le i\le \frac{t}{3}-1. Observe that P1=u1v1v2⋯vt−1vt{P}_{1}={u}_{1}{v}_{1}{v}_{2}\cdots {v}_{t-1}{v}_{t}and P2=u2v1vtvt−1⋯v3v2{P}_{2}={u}_{2}{v}_{1}{v}_{t}{v}_{t-1}\cdots {v}_{3}{v}_{2}are two total proper paths.Case 2. t≡1(mod3)t\equiv 1\left({\rm{mod}}\hspace{0.33em}3). Assign a total coloring ccto GGas follows: Let c(u1v1)=c(vt−1v1)=1,c(vt)=2,c(u2v1)=c(vtv1)=3c\left({u}_{1}{v}_{1})=c\left({v}_{t-1}{v}_{1})=1,c\left({v}_{t})=2,c\left({u}_{2}{v}_{1})=c\left({v}_{t}{v}_{1})=\hspace{-0.08em}3. Let iibe an integer with 0≤i≤t3−10\le i\le &#x230A;\hspace{-0.16em},\frac{t}{3},\hspace{-0.16em}&#x230B;-1, c(v3i+1)=c(v3i+2v3i+3)=2c\left({v}_{3i+1})=c\left({v}_{3i+2}{v}_{3i+3})=2, c(v3i+2)=c(v3i+3v3i+4)=1c\left({v}_{3i+2})=c\left({v}_{3i+3}{v}_{3i+4})=1, and c(v3i+3)=c(v3i+1u3i+2)=3c\left({v}_{3i+3})=c\left({v}_{3i+1}{u}_{3i+2})=3. We can find that P1=u1v1v2⋯vt−1vt{P}_{1}={u}_{1}{v}_{1}{v}_{2}\cdots {v}_{t-1}{v}_{t}and P2=u2v1vt−1⋯v3v2vt{P}_{2}={u}_{2}{v}_{1}{v}_{t-1}\cdots {v}_{3}{v}_{2}{v}_{t}are two total proper paths.Case 3. t≡2(mod3)t\equiv 2\left({\rm{mod}}\hspace{0.33em}3). Assign a total coloring ccto GGas follows: Let c(u1v1)=c(vtv1)=c(vt)=c(vt−2v1)=1,c(vt−1)=2,c(u2v1)=c(vtvt−1)=c(vt−1v2)=3c\left({u}_{1}{v}_{1})=c\left({v}_{t}{v}_{1})=c\left({v}_{t})=c\left({v}_{t-2}{v}_{1})=1,c\left({v}_{t-1})=2,c\left({u}_{2}{v}_{1})=c\left({v}_{t}{v}_{t-1})=c\left({v}_{t-1}{v}_{2})=3. Let iibe an integer with 0≤i≤t3−10\le i\le &#x230A;\hspace{-0.16em},\frac{t}{3},\hspace{-0.16em}&#x230B;-1, c(v3i+1)=c(v3i+2v3i+3)=2c\left({v}_{3i+1})=c\left({v}_{3i+2}{v}_{3i+3})=2, c(v3i+2)=c(v3i+3v3i+4)=1c\left({v}_{3i+2})=c\left({v}_{3i+3}{v}_{3i+4})=1, and c(v3i+3)=c(v3i+1u3i+2)=3c\left({v}_{3i+3})=c\left({v}_{3i+1}{u}_{3i+2})=3. We can easily verify that P1=u1v1v2⋯vt−1vt,P2=u2v1vt−2vt−3⋯v3v2vt−1{P}_{1}={u}_{1}{v}_{1}{v}_{2}\cdots {v}_{t-1}{v}_{t},{P}_{2}={u}_{2}{v}_{1}{v}_{t-2}{v}_{t-3}\cdots {v}_{3}{v}_{2}{v}_{t-1}, and P3=vtv1u2{P}_{3}={v}_{t}{v}_{1}{u}_{2}are three total proper paths.Thus, GGis total proper connected with the above coloring, and so tpc(G)=3{\rm{tpc}}\left(G)=3. This completes the proof of Claim 1.□Suppose F≅2K1F\cong 2{K}_{1}. Assume that NX(u1)∩NX(u2)=∅{N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{2})=\varnothing . Then GGhas a Hamiltonian path, and so tpc(G)=3{\rm{tpc}}\left(G)=3. Otherwise, if ∣NX(u1)∩NX(u2)∣=dX(u1)=dX(u2)=1| {N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{2})| ={d}_{X}\left({u}_{1})={d}_{X}\left({u}_{2})=1, then we know that tpc(G)=3{\rm{tpc}}\left(G)=3from Claim 1. If ∣NX(u1)∩NX(u2)∣≥2| {N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{2})| \ge 2, or ∣NX(u1)∩NX(u2)∣=1| {N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{2})| =1, and max{dX(u1),dX(u2)}≥2{\rm{\max }}\left\{{d}_{X}\left({u}_{1}),{d}_{X}\left({u}_{2})\right\}\ge 2, then we can find that GGhas a Hamiltonian path. Thus, tpc(G)=3{\rm{tpc}}\left(G)=3.5Proof of Theorem 1.4Suppose F≅K3F\cong {K}_{3}or P3{P}_{3}. Note that GGhas a Hamiltonian path, and so tpc(G)=3{\rm{tpc}}\left(G)=3. The following three claims will be used later.Claim 2. Let GGbe a graph obtained by adding a pendant vertex u3{u}_{3}adjacent to vertex u1{u}_{1}or u2{u}_{2}of graph in Claim 1. Then tpc(G)=3{\rm{tpc}}\left(G)=3.ProofWithout loss of generality, assume that u3{u}_{3}is adjacent to u1{u}_{1}. Let c(u1)={1,2,3}\{c(u1v1),c(v1)},c(u1u3)=c(v1)c\left({u}_{1})\left=\left\{1,2,3\right\}\backslash \left\{c\left({u}_{1}{v}_{1}),c\left({v}_{1})\right\}\right,c\left({u}_{1}{u}_{3})=c\left({v}_{1}), and the remaining vertices and edges are assigned the same color as Claim 1. We can verify that GGis total proper connected with the above coloring. Then tpc(G)=3{\rm{tpc}}\left(G)=3. This completes the proof of Claim 2.□Claim 3. Let GGbe a graph obtained by adding three vertices {u1,u2,u3}\left\{{u}_{1},{u}_{2},{u}_{3}\right\}to a complete graph Kt{K}_{t}such that d(u1)=d(u3)=1,d(u2)=2,N(u1)∩N(u3)=∅d\left({u}_{1})=d\left({u}_{3})=1,d\left({u}_{2})=2,N\left({u}_{1})\cap N\left({u}_{3})=\varnothing , and ∣N(u2)∩N(ui)∣=1| N\left({u}_{2})\cap N\left({u}_{i})| =1, where 1≤i≤21\le i\le 2. Then tpc(G)=3{\rm{tpc}}\left(G)=3.ProofWithout loss of generality, assume that N(u2)∩N(u1)=v1,N(u2)∩N(u3)=viN\left({u}_{2})\cap N\left({u}_{1})={v}_{1},N\left({u}_{2})\cap N\left({u}_{3})={v}_{i}. Let c(u3vi)=c(vivi+1),c(v1u2)=c(vivi−1)c\left({u}_{3}{v}_{i})=c\left({v}_{i}{v}_{i+1}),c\left({v}_{1}{u}_{2})=c\left({v}_{i}{v}_{i-1}), and the remaining vertices and edges are assigned the same color as Claim 1. Suppose t≡0(mod3)t\equiv 0\left({\rm{mod}}\hspace{0.33em}3). Note that P1=u1v1v2⋯vt−1vt{P}_{1}={u}_{1}{v}_{1}{v}_{2}\cdots {v}_{t-1}{v}_{t}, P2=u2v1vtvt−1⋯v3v2{P}_{2}={u}_{2}{v}_{1}{v}_{t}{v}_{t-1}\cdots {v}_{3}{v}_{2}, P3=u3viu2{P}_{3}={u}_{3}{v}_{i}{u}_{2}, P4=u1v1v2⋯viu3{P}_{4}={u}_{1}{v}_{1}{v}_{2}\cdots {v}_{i}{u}_{3}, and P5=u3vivi−1⋯v1vtvt−1⋯vi+1{P}_{5}={u}_{3}{v}_{i}{v}_{i-1}\cdots {v}_{1}{v}_{t}{v}_{t-1}\cdots {v}_{i+1}are five total proper paths. Suppose t≡1(mod3)t\equiv 1\left({\rm{mod}}\hspace{0.33em}3). Note that P1=u1v1v2⋯vt−1vt{P}_{1}={u}_{1}{v}_{1}{v}_{2}\cdots {v}_{t-1}{v}_{t}, P2=u2v1vt−1⋯v3v2vt{P}_{2}={u}_{2}{v}_{1}{v}_{t-1}\cdots {v}_{3}{v}_{2}{v}_{t}, P3=u3viu2{P}_{3}={u}_{3}{v}_{i}{u}_{2}, P4=u1v1v2⋯viu3{P}_{4}={u}_{1}{v}_{1}{v}_{2}\cdots {v}_{i}{u}_{3}and P5=u3vivi−1⋯v2vtvt−1⋯vi+1{P}_{5}={u}_{3}{v}_{i}{v}_{i-1}\cdots {v}_{2}{v}_{t}{v}_{t-1}\cdots {v}_{i+1}are five total proper paths. Suppose t≡2(mod3)t\hspace{-0.08em}\equiv \hspace{-0.08em}2\left({\rm{mod}}\hspace{0.25em}3). We can find that P1=u1v1v2⋯vt−1vt{P}_{1}\hspace{-0.08em}=\hspace{-0.08em}{u}_{1}{v}_{1}{v}_{2}\cdots {v}_{t-1}{v}_{t}, P2=u2v1vt−2⋯v3v2vt−1{P}_{2}={u}_{2}{v}_{1}{v}_{t-2}\cdots {v}_{3}{v}_{2}{v}_{t-1}, P3=u3viu2{P}_{3}={u}_{3}{v}_{i}{u}_{2}, P4=u1v1v2⋯viu3{P}_{4}={u}_{1}{v}_{1}{v}_{2}\cdots {v}_{i}{u}_{3}, P5=u3vivi−1⋯v2vt−1vt−2⋯vi+1{P}_{5}={u}_{3}{v}_{i}{v}_{i-1}\cdots {v}_{2}{v}_{t-1}{v}_{t-2}\cdots {v}_{i+1}, and P6=u3vivi−1vt{P}_{6}={u}_{3}{v}_{i}{v}_{i-1}{v}_{t}are six total proper paths. Therefore, GGis total proper connected with the above coloring, and so tpc(G)=3{\rm{tpc}}\left(G)=3. This completes the proof of Claim 3.□Claim 4. Let GGbe a graph obtained by adding a vertex uuto the graph in Claim 1 such that d(u)=2d\left(u)=2and v1{v}_{1}is adjacent to uu. Then tpc(G)=3{\rm{tpc}}\left(G)=3.ProofSince d(u)=2d\left(u)=2, without loss of generality, we assume that vi{v}_{i}is adjacent to uu. Let c(uv1)=1,c(uvi)={1,2,3}\{c(vi),c(vivi−1)}c\left(u{v}_{1})=1,c\left(u{v}_{i})\left=\left\{1,2,3\right\}\backslash \left\{c\left({v}_{i}),c\left({v}_{i}{v}_{i-1})\right\}, and the remaining vertices and edges are assigned the same color as Claim 1. Suppose t≡0(mod3)t\equiv 0\left({\rm{mod}}\hspace{0.33em}3). Note that P1=uvivi−1⋯v1vtvt−1⋯vi+1,P2=uvivi−1⋯v1u1{P}_{1}=u{v}_{i}{v}_{i-1}\cdots {v}_{1}{v}_{t}{v}_{t-1}\cdots {v}_{i+1},{P}_{2}=u{v}_{i}{v}_{i-1}\cdots {v}_{1}{u}_{1}and P3=uv1u2{P}_{3}=u{v}_{1}{u}_{2}are three total proper paths. Suppose t≡1(mod3)t\equiv 1\left({\rm{mod}}\hspace{0.33em}3). Note that P1=uvivi−1⋯v1u1{P}_{1}=u{v}_{i}{v}_{i-1}\cdots {v}_{1}{u}_{1}, P2=uvivi−1⋯v2vtvt−1⋯vi+1{P}_{2}=u{v}_{i}{v}_{i-1}\cdots {v}_{2}{v}_{t}{v}_{t-1}\cdots {v}_{i+1}, and P3=uv1u2{P}_{3}=u{v}_{1}{u}_{2}are three total proper paths. Suppose t≡2(mod3)t\equiv 2\left({\rm{mod}}\hspace{0.33em}3). We can find that P1=uvivi−1⋯v1u1{P}_{1}=u{v}_{i}{v}_{i-1}\cdots {v}_{1}{u}_{1}, P2=uvivi−1⋯v2vt−1vt−2⋯vi+1{P}_{2}=u{v}_{i}{v}_{i-1}\cdots {v}_{2}{v}_{t-1}{v}_{t-2}\cdots {v}_{i+1}, P3=uv1u2{P}_{3}=u{v}_{1}{u}_{2}, and P4=uvivi−1⋯v1vt{P}_{4}=u{v}_{i}{v}_{i-1}\cdots {v}_{1}{v}_{t}are four total proper paths. Hence, GGis total proper connected with the above coloring, and so tpc(G)=3{\rm{tpc}}\left(G)=3. This completes the proof of Claim 4.□Suppose F≅K2+K1F\cong {K}_{2}+{K}_{1}. Let V(K2)={u1,u2}V\left({K}_{2})=\left\{{u}_{1},{u}_{2}\right\}and V(K1)={u3}V\left({K}_{1})=\left\{{u}_{3}\right\}. Since diam(G)=2{\rm{diam}}\left(G)=2, we have NX(u1)∩NX(u2)∩NX(u3)={v}{N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{2})\cap {N}_{X}\left({u}_{3})=\left\{v\right\}, and so tpc(G)=3{\rm{tpc}}\left(G)=3by Claim 2. Suppose F≅3K1F\cong 3{K}_{1}. Assume NX(u1)∩NX(u2)∩NX(u3)=∅{N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{2})\cap {N}_{X}\left({u}_{3})=\varnothing . Then tpc(G)=3{\rm{tpc}}\left(G)=3by Claim 3. Assume NX(u1)∩NX(u2)∩NX(u3)≠∅{N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{2})\cap {N}_{X}\left({u}_{3})\ne \varnothing . If dX(u1)=dX(u2)=dX(u3)=1{d}_{X}\left({u}_{1})={d}_{X}\left({u}_{2})={d}_{X}\left({u}_{3})=1, then tpc(G)≥4{\rm{tpc}}\left(G)\ge 4by [26, Proposition 2]. Define a total coloring of GGas follows: c(u3v)=4c\left({u}_{3}v)=4with v∈NX(u3)v\in {N}_{X}\left({u}_{3}), and the remaining vertices and edges are assigned the same color as Claim 1. We check that any two vertices have a total proper path, and so tpc(G)=4{\rm{tpc}}\left(G)=4. Otherwise, we have dX(u1)+dX(u2)+dX(u3)≥4{d}_{X}\left({u}_{1})+{d}_{X}\left({u}_{2})+{d}_{X}\left({u}_{3})\ge 4. Without loss of generality, let dX(u1)≥2{d}_{X}\left({u}_{1})\ge 2, and u∈X\{v}u\left\in X\backslash \left\{v\right\}where v∈NX(u1)∩NX(u2)∩NX(u3)v\in {N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{2})\cap {N}_{X}\left({u}_{3}). Thus, tpc(G)=3{\rm{tpc}}\left(G)=3by Claim 4.6Proof of Theorem 1.5Case 1. diam(G)=3{\rm{diam}}\left(G)=3. We prove Case 1 by analyzing the structure of FF.Subcase 1.1. F≅K3F\cong {K}_{3}or P3{P}_{3}. Note that GGhas a Hamiltonian path, and so tpc(G)=3{\rm{tpc}}\left(G)=3.Subcase 1.2. F≅K2+K1F\cong {K}_{2}+{K}_{1}. Denote V(K2)={u1,u2}V\left({K}_{2})=\left\{{u}_{1},{u}_{2}\right\}and V(K1)={u3}V\left({K}_{1})=\left\{{u}_{3}\right\}. Suppose NX(u1)∩NX(u3)≠∅{N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{3})\ne \varnothing or NX(u2)∩NX(u3)≠∅{N}_{X}\left({u}_{2})\cap {N}_{X}\left({u}_{3})\ne \varnothing . Without of loss generality, we may assume that NX(u1)∩NX(u3)≠∅{N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{3})\ne \varnothing . Then tpc(G)=3{\rm{tpc}}\left(G)=3by Claim 2. Suppose NX(u1)∩NX(u3)=∅{N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{3})=\varnothing and NX(u2)∩NX(u3)=∅{N}_{X}\left({u}_{2})\cap {N}_{X}\left({u}_{3})=\varnothing . Since diam(G)=3{\rm{diam}}\left(G)=3, we have NX(u1)≠∅{N}_{X}\left({u}_{1})\ne \varnothing and NX(u3)≠∅{N}_{X}\left({u}_{3})\ne \varnothing . Then GGhas a Hamiltonian path, and so tpc(G)=3{\rm{tpc}}\left(G)=3.Subcase 1.3. F≅3K1F\cong 3{K}_{1}. Let V(F)={u1,u2,u3}V\left(F)=\left\{{u}_{1},{u}_{2},{u}_{3}\right\}. Since diam(G)=3{\rm{diam}}\left(G)=3, we have NX(u1)∩NX(u2)∩NX(u3)=∅{N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{2})\cap {N}_{X}\left({u}_{3})=\varnothing . Suppose there exists two vertices ui,uj∈V(F){u}_{i},{u}_{j}\in V\left(F)satisfy NX(ui)∩NX(uj)≠∅{N}_{X}\left({u}_{i})\cap {N}_{X}\left({u}_{j})\ne \varnothing . Without loss of generality, let u1{u}_{1}and u2{u}_{2}satisfy NX(u1)∩NX(u2)≠∅{N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{2})\ne \varnothing and v1∈NX(u1)∩NX(u2){v}_{1}\in {N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{2}). Assume dX(u1)=dX(u2)=1{d}_{X}\left({u}_{1})={d}_{X}\left({u}_{2})=1and vi∈N(u3){v}_{i}\in N\left({u}_{3}). Since GGis not complete, we have tpc(G)≥3{\rm{tpc}}\left(G)\ge 3. To the contrary, suppose there exists a total coloring ccof GGusing three colors such that GGis total proper connected. Since any two vertices of GGare connected by a total proper path, we have c(u1v1)≠c(v1)≠c(u2v1)c\left({u}_{1}{v}_{1})\ne c\left({v}_{1})\ne c\left({u}_{2}{v}_{1}). Without loss of generality, let c(u1v1)=1,c(v1)=2c\left({u}_{1}{v}_{1})=1,c\left({v}_{1})=2and c(u2v1)=3c\left({u}_{2}{v}_{1})=3. Consider the total proper path PPbetween u1{u}_{1}and uu, then the color of vertices and edges in PPfollows the sequence 1,2,3,…,1,2,3,…1,2,3,\ldots ,1,2,3,\ldots . Thus, the value of (c(vi),c(viu))\left(c\left({v}_{i}),c\left({v}_{i}u))is (1,2),(2,3)\left(1,2),\left(2,3), or (3,1)\left(3,1). Consider the total proper path QQbetween u2{u}_{2}and uu, then the color of vertices and edges in QQfollows the sequence 3,2,1,…,3,2,1,…3,2,1,\ldots ,3,2,1,\ldots . But the value of (c(vi),c(viu))\left(c\left({v}_{i}),c\left({v}_{i}u))is (3,2),(2,1)\left(3,2),\left(2,1), or (1,3)\left(1,3), a contradiction. Assign a total coloring ccto GGas follows: c(u1v1)=1,c(v1)=c(viu)=2,c(u2v1)=3c\left({u}_{1}{v}_{1})=1,c\left({v}_{1})=c\left({v}_{i}u)=2,c\left({u}_{2}{v}_{1})=3, assign 4 to the remaining edges, and assign 1 to the remaining vertices. We can check that GGis total proper connected with the above coloring, and so tpc(G)=4{\rm{tpc}}\left(G)=4. Assume dX(u1)+dX(u2)≥3{d}_{X}\left({u}_{1})+{d}_{X}\left({u}_{2})\ge 3, without loss of generality, let dX(u1)≥2{d}_{X}\left({u}_{1})\ge 2. If dX(u3)=1{d}_{X}\left({u}_{3})=1and NX(u1)∩NX(u3)≠∅{N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{3})\ne \varnothing , then we have tpc(G)=3{\rm{tpc}}\left(G)=3by Claim 3; if NX(u1)∩NX(u3)=∅{N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{3})=\varnothing , or NX(u1)∩NX(u3)≠∅{N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{3})\ne \varnothing and dX(u3)≥2{d}_{X}\left({u}_{3})\ge 2, then GGhas a Hamiltonian path, and so tpc(G)=3{\rm{tpc}}\left(G)=3.Now, we may suppose NX(u1)∩NX(u2)=∅,NX(u1)∩NX(u3)=∅{N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{2})=\varnothing ,{N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{3})=\varnothing , and NX(u2)∩NX(u3)=∅{N}_{X}\left({u}_{2})\cap {N}_{X}\left({u}_{3})=\varnothing . Since GGis not complete, we have tpc(G)≥3{\rm{tpc}}\left(G)\ge 3. To the contrary, assume that vi∈N(ui){v}_{i}\in N\left({u}_{i}), and there exists a total coloring ccof GGusing three colors such that GGis total proper connected. Since any two vertices of GGare connected by a total proper path, we have c(u1v1)≠c(v1)c\left({u}_{1}{v}_{1})\ne c\left({v}_{1}). Without loss of generality, let c(u1v1)=1c\left({u}_{1}{v}_{1})=1and c(v1)=2c\left({v}_{1})=2. Consider the total proper path PPbetween u1{u}_{1}and u2{u}_{2}, then the color of vertices and edges in PPfollows the sequence 1,2,3,…,1,2,3,…1,2,3,\ldots ,1,2,3,\ldots . Thus, the value of (c(v2),c(v2u2))\left(c\left({v}_{2}),c\left({v}_{2}{u}_{2}))is (1,2),(2,3)\left(1,2),\left(2,3), or (3,1)\left(3,1). Consider the total proper path QQbetween u1{u}_{1}and u3{u}_{3}, then the color of vertices and edges in QQfollows the sequence 1,2,3,…,1,2,3,…1,2,3,\ldots ,1,2,3,\ldots . Hence, the value of (c(v3),c(v3u3))\left(c\left({v}_{3}),c\left({v}_{3}{u}_{3}))is (1,2),(2,3)\left(1,2),\left(2,3), or (3,1)\left(3,1). Consider the total proper path WWbetween u2{u}_{2}and u3{u}_{3}. If c(v2)=1c\left({v}_{2})=1and c(v2u2)=2c\left({v}_{2}{u}_{2})=2, then the color of vertices and edges in WWfollows the sequence 2,1,3,…,2,1,3,…2,1,3,\ldots ,2,1,3,\ldots . Note that the value of (c(v3),c(v3u3))\left(c\left({v}_{3}),c\left({v}_{3}{u}_{3}))is (2,1),(1,3)\left(2,1),\left(1,3), or (3,2)\left(3,2), a contradiction. If c(v2)=2c\left({v}_{2})=2and c(v2u2)=3c\left({v}_{2}{u}_{2})=3, then the color of vertices and edges in WWfollows the sequence 3,2,1,…,3,2,1,…3,2,1,\ldots ,3,2,1,\ldots . Note that the value of (c(v3),c(v3u3))\left(c\left({v}_{3}),c\left({v}_{3}{u}_{3}))is (2,1),(1,3)\left(2,1),\left(1,3), or (3,2)\left(3,2), a contradiction. If c(v2)=3c\left({v}_{2})=3and c(v2u2)=1c\left({v}_{2}{u}_{2})=1, then the color of vertices and edges in WWfollows the sequence 1,3,2,…,1,3,2,…1,3,2,\ldots ,1,3,2,\ldots . Note that the value of (c(v3),c(v3u3))\left(c\left({v}_{3}),c\left({v}_{3}{u}_{3}))is (2,1),(1,3)\left(2,1),\left(1,3), or (3,2)\left(3,2), a contradiction. Assign a total coloring ccto GGas follows: c(u1v1)=c(v2)=1,c(v1)=c(u2v2)=c(u3v3)=2c\left({u}_{1}{v}_{1})=c\left({v}_{2})=1,c\left({v}_{1})=c\left({u}_{2}{v}_{2})=c\left({u}_{3}{v}_{3})=2, assign 4 to the remaining vertices, and assign 3 to the remaining edges. We can verify that GGis total proper connected with the above coloring, and so tpc(G)=4{\rm{tpc}}\left(G)=4.Case 2. diam(G)≥4.{\rm{diam}}\left(G)\ge 4.Thus, F≅P3F\cong {P}_{3}or F≅K2+K1F\cong {K}_{2}+{K}_{1}. Assume F≅P3F\cong {P}_{3}. Obviously, GGhas a Hamiltonian path, and so tpc(G)=3{\rm{tpc}}\left(G)=3. Assume F≅K2+K1F\cong {K}_{2}+{K}_{1}. Denote V(K2)={u1,u2}V\left({K}_{2})=\left\{{u}_{1},{u}_{2}\right\}and V(K1)={u3}V\left({K}_{1})=\left\{{u}_{3}\right\}, without loss of generality, we have dX(u2)=0,dX(u1)≥1,dX(u3)≥1{d}_{X}\left({u}_{2})=0,{d}_{X}\left({u}_{1})\ge 1,{d}_{X}\left({u}_{3})\ge 1satisfying NX(u1)∩NX(u2)=∅{N}_{X}\left({u}_{1})\cap {N}_{X}\left({u}_{2})=\varnothing . Hence, we can find that GGhas a Hamiltonian path, and so tpc(G)=3{\rm{tpc}}\left(G)=3.7Proof of Theorem 1.6The proof of Theorem 1.6 follows from the next two lemmas. First, we shall determine the total proper kk-connection numbers of the circular ladders.Lemma 7.1Let nnbe an integer with n≥3n\ge 3. Then tpc(CL2n)=tpc2(CL2n)=3,tpc3(CL2n)=4{\rm{tpc}}\left({{\rm{CL}}}_{2n})={{\rm{tpc}}}_{2}\left({{\rm{CL}}}_{2n})=3,{{\rm{tpc}}}_{3}\left({{\rm{CL}}}_{2n})=4.ProofLet nnbe an integer with n≥3n\ge 3. Since CL2n{{\rm{CL}}}_{2n}contains a Hamiltonian path that is not complete, we have tpc(CL2n)=3{\rm{tpc}}\left({{\rm{CL}}}_{2n})=3. Since tpc2(CL2n)≥tpc(CL2n)=3{{\rm{tpc}}}_{2}\left({{\rm{CL}}}_{2n})\ge {\rm{tpc}}\left({{\rm{CL}}}_{2n})=3, we only need to prove tpc2(CL2n)≤3{{\rm{tpc}}}_{2}\left({{\rm{CL}}}_{2n})\le 3.Case 1. n≡0(mod3)n\equiv 0\left({\rm{mod}}\hspace{0.33em}3). Let n=3tn=3t. Assign a total coloring ccto CL2n{{\rm{CL}}}_{2n}as follows: Let iibe an integer with 0≤i≤t−10\le i\le t\hspace{-0.08em}-\hspace{-0.08em}1, c(u3i+1)=c(u3i+2u3i+3)=c(v3i+2)=c(v3i+3v3i+4)=1c\left({u}_{3i+1})\hspace{-0.08em}=\hspace{-0.08em}c\left({u}_{3i+2}{u}_{3i+3})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{3i+2})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{3i+3}{v}_{3i+4})=1, c(u3i+2)=c(u3i+3u3i+4)=c(v3i+3)=c(v3i+1v3i+2)=2c\left({u}_{3i+2})=c\left({u}_{3i+3}{u}_{3i+4})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{3i+3})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{3i+1}{v}_{3i+2})\hspace{-0.08em}=\hspace{-0.08em}2, and c(u3i+3)=c(u3i+1u3i+2)=c(v3i+1)=c(v3i+2v3i+3)=3c\left({u}_{3i+3})=c\left({u}_{3i+1}{u}_{3i+2})=c\left({v}_{3i+1})=c\left({v}_{3i+2}{v}_{3i+3})=3; c(uivi),c(ui),c(vi)∈{1,2,3}c\left({u}_{i}{v}_{i}),c\left({u}_{i}),c\left({v}_{i})\in \left\{1,2,3\right\}with c(uivi)≠c(ui)≠c(vi)c\left({u}_{i}{v}_{i})\ne c\left({u}_{i})\ne c\left({v}_{i})for 1≤i≤n1\le i\le n. Let xxand yybe any two distinct vertices of CL2n{{\rm{CL}}}_{2n}. By symmetry, we may assume that x=u1x={u}_{1}. If y=uiy={u}_{i}for 2≤i≤n2\le i\le n, then xu2u3⋯ui−1yx{u}_{2}{u}_{3}\cdots {u}_{i-1}yand xunun−1⋯ui+1yx{u}_{n}{u}_{n-1}\cdots {u}_{i+1}yare two total proper paths connecting xxand yy. If y=v1y={v}_{1}, then xyxyand xunun−1⋯u2yx{u}_{n}{u}_{n-1}\cdots {u}_{2}yare two total proper paths connecting xxand yy. If y=viy={v}_{i}, then xv1vnvn−1⋯vi+1yx{v}_{1}{v}_{n}{v}_{n-1}\cdots {v}_{i+1}yand xunun−1⋯uiyx{u}_{n}{u}_{n-1}\cdots {u}_{i}yare two total proper paths connecting xxand yy. Thus, CL2n{{\rm{CL}}}_{2n}is total proper 2-connected with the above coloring.Case 2. n≡1(mod3)n\equiv 1\left({\rm{mod}}\hspace{0.33em}3). Let n=3t+1n=3t+1. Define a total coloring ccof CL2n{{\rm{CL}}}_{2n}as follows: Let iibe an integer with 0≤i≤t−10\le i\le t\hspace{-0.08em}-\hspace{-0.08em}1, c(u3i+1)=c(u3i+2u3i+3)=c(v3i+3)=c(v3i+1v3i+2)=1c\left({u}_{3i+1})\hspace{-0.08em}=\hspace{-0.08em}c\left({u}_{3i+2}{u}_{3i+3})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{3i+3})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{3i+1}{v}_{3i+2})\hspace{-0.08em}=\hspace{-0.08em}1, c(u3i+2)=c(u3i+3u3i+4)=c(v3i+2)=c(v3i+3v3i+4)=3c\left({u}_{3i+2})\hspace{-0.08em}=\hspace{-0.08em}c\left({u}_{3i+3}{u}_{3i+4})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{3i+2})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{3i+3}{v}_{3i+4})=3, and c(u3i+3)=c(u3i+1u3i+2)=c(v3i+1)=c(v3i+2v3i+3)=2c\left({u}_{3i+3})=c\left({u}_{3i+1}{u}_{3i+2})=c\left({v}_{3i+1})=c\left({v}_{3i+2}{v}_{3i+3})=2; c(un)=c(v1vn)=1,c(unvn)=2,c(vn)=c(u1un)=3c\left({u}_{n})=c\left({v}_{1}{v}_{n})=1,c\left({u}_{n}{v}_{n})=2,c\left({v}_{n})=c\left({u}_{1}{u}_{n})=3, and c(ujvj)=3c\left({u}_{j}{v}_{j})=3for 1≤j≤n−11\le j\le n-1. Let xxand yybe any two distinct vertices of CL2n{{\rm{CL}}}_{2n}. We may assume that x=uix={u}_{i}for 1≤i≤n−11\le i\le n-1. If y=ujy={u}_{j}for i≤j≤n−1i\le j\le n-1, then xui+1ui+2⋯uj−1yx{u}_{i+1}{u}_{i+2}\cdots {u}_{j-1}yand xui−1ui−2⋯u1v1v2⋯vn−1un−1un−2⋯uj+1yx{u}_{i-1}{u}_{i-2}\cdots {u}_{1}{v}_{1}{v}_{2}\cdots {v}_{n-1}{u}_{n-1}{u}_{n-2}\cdots {u}_{j+1}yare two total proper paths connecting xxand yy. If y=vjy={v}_{j}for 1≤j≤n−11\le j\le n-1, then xui−1ui−2⋯u1v1v2⋯vj−1yx{u}_{i-1}{u}_{i-2}\cdots {u}_{1}{v}_{1}{v}_{2}\cdots {v}_{j-1}yand xui+1⋯un−1vn−1vn−2⋯vj+1yx{u}_{i+1}\cdots {u}_{n-1}{v}_{n-1}{v}_{n-2}\cdots {v}_{j+1}yare two total proper paths connecting xxand yy. If y=uny={u}_{n}, then xui+1⋯un−1yx{u}_{i+1}\cdots {u}_{n-1}yand xui−1ui−2⋯u1yx{u}_{i-1}{u}_{i-2}\cdots {u}_{1}yare two total proper paths connecting xxand yy. If y=vny={v}_{n}, then xui+1⋯un−1unyx{u}_{i+1}\cdots {u}_{n-1}{u}_{n}yand xui−1ui−2⋯u1v1v2⋯vn−1yx{u}_{i-1}{u}_{i-2}\cdots {u}_{1}{v}_{1}{v}_{2}\cdots {v}_{n-1}yare two total proper paths connecting xxand yy. Assume that x=unx={u}_{n}. If y=vjy={v}_{j}for 1≤j≤n−11\le j\le n-1, then xu1u2⋯un−1vn−1vn−2⋯vj+1yx{u}_{1}{u}_{2}\cdots {u}_{n-1}{v}_{n-1}{v}_{n-2}\cdots {v}_{j+1}yand xunvnv1v2⋯vj−1yx{u}_{n}{v}_{n}{v}_{1}{v}_{2}\cdots {v}_{j-1}yare two total proper paths connecting xxand yy. If y=vny={v}_{n}, then xyxyand xun−1un−2⋯u1v1yx{u}_{n-1}{u}_{n-2}\cdots {u}_{1}{v}_{1}yare two total proper paths connecting xxand yy. Thus, CL2n{{\rm{CL}}}_{2n}is total proper 2-connected with the above coloring.Case 3. n≡2(mod3)n\equiv 2\left({\rm{mod}}\hspace{0.33em}3). Let n=3t+2n=3t+2. Define a total coloring ccof CL2n{{\rm{CL}}}_{2n}as follows: Let iibe an integer with 0≤i≤t−20\le i\le t\hspace{-0.08em}-\hspace{-0.08em}2, c(u3i+1)=c(u3i+2u3i+3)=c(v3i+1)=c(v3i+2v3i+3)=1c\left({u}_{3i+1})\hspace{-0.08em}=\hspace{-0.08em}c\left({u}_{3i+2}{u}_{3i+3})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{3i+1})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{3i+2}{v}_{3i+3})\hspace{-0.08em}=\hspace{-0.08em}1, c(u3i+2)=c(u3i+3u3i+4)=c(v3i+2)=c(v3i+3v3i+4)=3c\left({u}_{3i+2})\hspace{-0.08em}=\hspace{-0.08em}c\left({u}_{3i+3}{u}_{3i+4})=c\left({v}_{3i+2})=c\left({v}_{3i+3}{v}_{3i+4})=3, and c(u3i+3)=c(u3i+1u3i+2)=c(v3i+1)=c(v3i+2v3i+3)=2c\left({u}_{3i+3})\hspace{-0.08em}=\hspace{-0.08em}c\left({u}_{3i+1}{u}_{3i+2})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{3i+1})=c\left({v}_{3i+2}{v}_{3i+3})=2; c(un)=c(un−4)=c(vn−2)=c(vn)=1c\left({u}_{n})=c\left({u}_{n-4})=c\left({v}_{n-2})=c\left({v}_{n})=1, c(un−1)=c\left({u}_{n-1})=c(vn−1)=c(un−3)=c(vn−3)=3c\left({v}_{n-1})\hspace{0.12em}=\hspace{0.12em}c\left({u}_{n-3})\hspace{0.12em}=\hspace{0.12em}c\left({v}_{n-3})\hspace{0.12em}=\hspace{0.12em}3, c(un−2)=c(vn−4)=2c\left({u}_{n-2})\hspace{0.12em}=\hspace{0.12em}c\left({v}_{n-4})\hspace{0.12em}=\hspace{0.12em}2, c(un−4un−3)=c(un−1un)=c(vn−1vn)=c(vn−3vn−2)=c(vn−2vn−1)=2c\left({u}_{n-4}{u}_{n-3})\hspace{0.12em}=\hspace{0.12em}c\left({u}_{n-1}{u}_{n})\hspace{0.12em}=\hspace{0.12em}c\left({v}_{n-1}{v}_{n})\hspace{0.12em}=\hspace{0.12em}c\left({v}_{n-3}{v}_{n-2})=c({v}_{n-2}{v}_{n-1})=2, c(un−3un−2)=c(un−2un−1)=c(vn−4vn−3)=c(u2v2)=1c\left({u}_{n-3}{u}_{n-2})=c\left({u}_{n-2}{u}_{n-1})=c\left({v}_{n-4}{v}_{n-3})=c\left({u}_{2}{v}_{2})=1, c(u1u1)=c(u1un)=c(v1vn)=3c\left({u}_{1}{u}_{1})=c\left({u}_{1}{u}_{n})=c\left({v}_{1}{v}_{n})=3; c(u4jv4j)=1c\left({u}_{4j}{v}_{4j})=1, c(u4j−1v4j−1)=2c\left({u}_{4j-1}{v}_{4j-1})=2, c(u4j+1v4j+1)=3c\left({u}_{4j+1}{v}_{4j+1})=3, where 1≤j≤t−11\le j\le t-1. Note that u1u2⋯un−2vn−2vn−3⋯v1u1{u}_{1}{u}_{2}\cdots {u}_{n-2}{v}_{n-2}{v}_{n-3}\cdots {v}_{1}{u}_{1}is a total proper cycle, any two distinct vertices of the cycle have two disjoint total proper paths. Now, we may assume that x=uix={u}_{i}for 1≤i≤n−21\le i\le n-2. If y=un−1y={u}_{n-1}, then xui+1⋯un−2vn−2vn−1yx{u}_{i+1}\cdots {u}_{n-2}{v}_{n-2}{v}_{n-1}yand xui−1⋯u1unyx{u}_{i-1}\cdots {u}_{1}{u}_{n}yare two total proper paths connecting xxand yy. If y=uny={u}_{n}, then xui+1⋯un−2vn−2vn−1⋯v1vnyx{u}_{i+1}\cdots {u}_{n-2}{v}_{n-2}{v}_{n-1}\cdots {v}_{1}{v}_{n}yand xui−1⋯u1v1vnyx{u}_{i-1}\cdots {u}_{1}{v}_{1}{v}_{n}yare two total proper paths connecting xxand yy. If y=vn−1y={v}_{n-1}, then xui−1⋯u3v3v2v1vnyx{u}_{i-1}\cdots {u}_{3}{v}_{3}{v}_{2}{v}_{1}{v}_{n}yand xui+1⋯un−2vn−2yx{u}_{i+1}\cdots {u}_{n-2}{v}_{n-2}yare two total proper paths connecting xxand yy. If y=vny={v}_{n}, then xui−1⋯u1unyx{u}_{i-1}\cdots {u}_{1}{u}_{n}yand xui+1⋯un−2vn−2vn−1yx{u}_{i+1}\cdots {u}_{n-2}{v}_{n-2}{v}_{n-1}yare two total proper paths connecting xxand yy. Assume that x=vix={v}_{i}for 1≤i≤n−21\le i\le n-2. If y=vn−1y={v}_{n-1}, then xvi+1⋯vn−2un−2un−1⋯u1unun−1yx{v}_{i+1}\cdots {v}_{n-2}{u}_{n-2}{u}_{n-1}\cdots {u}_{1}{u}_{n}{u}_{n-1}yand xvi−1⋯v1vnyx{v}_{i-1}\cdots {v}_{1}{v}_{n}yare two total proper paths connecting xxand yy. If y=vny={v}_{n}, then xvi+1⋯vn−2un−2un−1unyx{v}_{i+1}\cdots {v}_{n-2}{u}_{n-2}{u}_{n-1}{u}_{n}yand xvi−1⋯v1yx{v}_{i-1}\cdots {v}_{1}yare two total proper paths connecting xxand yy. Thus, CL2n{{\rm{CL}}}_{2n}is total proper 2-connected with the above coloring.To the contrary, suppose there exists a total proper 3-connected coloring ccof CL2n{{\rm{CL}}}_{2n}using three colors. Considering u1{u}_{1}and v2{v}_{2}, u1u2v2,u1v1v2{u}_{1}{u}_{2}{v}_{2},{u}_{1}{v}_{1}{v}_{2}, and u1unvnvn−1⋯v3v2{u}_{1}{u}_{n}{v}_{n}{v}_{n-1}\cdots {v}_{3}{v}_{2}must be three total proper paths connecting u1{u}_{1}and v2{v}_{2}. Then c(u1u2)≠c(u2v2)≠c(u2)c\left({u}_{1}{u}_{2})\ne c\left({u}_{2}{v}_{2})\ne c\left({u}_{2}). Considering u1{u}_{1}and u3{u}_{3}, u1u2u3,u1unun−1⋯u4u3{u}_{1}{u}_{2}{u}_{3},{u}_{1}{u}_{n}{u}_{n-1}\cdots {u}_{4}{u}_{3}, and u1v1v2v3u3{u}_{1}{v}_{1}{v}_{2}{v}_{3}{u}_{3}must be three total proper paths connecting u1{u}_{1}and u3{u}_{3}. Hence, c(u1u2)≠c(u2u3)≠c(u2)c\left({u}_{1}{u}_{2})\ne c\left({u}_{2}{u}_{3})\ne c\left({u}_{2}), and so c(u2v2)=c(u2u3)c\left({u}_{2}{v}_{2})=c\left({u}_{2}{u}_{3}). But then, there is no set of three disjoint total proper paths connecting u3{u}_{3}and v2{v}_{2}, a contradiction. Hence, tpc3(CL2n)≥4{{\rm{tpc}}}_{3}\left({{\rm{CL}}}_{2n})\ge 4. Now we only need to prove tpc3(CL2n)≤4{{\rm{tpc}}}_{3}\left({{\rm{CL}}}_{2n})\le 4.Case 1. n≡0(mod3)n\equiv 0\left({\rm{mod}}\hspace{0.33em}3). Let n=3tn=3t. Assign a total coloring ccto CL2n{{\rm{CL}}}_{2n}as follows: Let iibe an integer with 0≤i≤t−10\hspace{-0.1em}\le i\hspace{-0.1em}\le \hspace{-0.1em}t\hspace{-0.1em}-\hspace{-0.1em}1, c(u3i+1)=c(u3i+2u3i+3)=c(v3i+2)=c(v3i+3v3i+4)=1c\left({u}_{3i+1})\hspace{-0.1em}=\hspace{-0.1em}c\left({u}_{3i+2}{u}_{3i+3})=c\left({v}_{3i+2})\hspace{-0.1em}=\hspace{-0.1em}c\left({v}_{3i+3}{v}_{3i+4})\hspace{-0.1em}=\hspace{-0.1em}1, c(u3i+2)=c(u3i+3u3i+4)=c(v3i+3)=c(v3i+1v3i+2)=2c\left({u}_{3i+2})\hspace{-0.1em}=\hspace{-0.1em}c\left({u}_{3i+3}{u}_{3i+4})\hspace{-0.1em}=\hspace{-0.1em}c\left({v}_{3i+3})\hspace{-0.1em}=\hspace{-0.1em}c\left({v}_{3i+1}{v}_{3i+2})\hspace{-0.1em}=\hspace{-0.1em}2, and c(u3i+3)=c(u3i+1u3i+2)=c(v3i+1)=c(v3i+2v3i+3)=3c\left({u}_{3i+3})=c\left({u}_{3i+1}{u}_{3i+2})=c\left({v}_{3i+1})=c\left({v}_{3i+2}{v}_{3i+3})=3; c(ujvj)=4c\left({u}_{j}{v}_{j})=4for 1≤j≤n1\le j\le n. Let xxand yybe any two distinct vertices of CL2n{{\rm{CL}}}_{2n}. By symmetry, we may assume that x=u1x={u}_{1}. If y=uiy={u}_{i}for 2≤i≤n2\le i\le n, then xu2u3⋯ui−1y,xunun−1⋯ui+1yx{u}_{2}{u}_{3}\cdots {u}_{i-1}y,x{u}_{n}{u}_{n-1}\cdots {u}_{i+1}y, and xv1v2⋯viyx{v}_{1}{v}_{2}\cdots {v}_{i}yare three total proper paths connecting xxand yy. If y=v1y={v}_{1}, then xy,xu2v2yxy,x{u}_{2}{v}_{2}yand xunvnyx{u}_{n}{v}_{n}yare three total proper paths connecting xxand yy. If y=viy={v}_{i}for 2≤i≤n2\le i\le n, then xu1u2⋯uiy,xv1v2⋯vi−1yx{u}_{1}{u}_{2}\cdots {u}_{i}y,x{v}_{1}{v}_{2}\cdots {v}_{i-1}y, and xunvnvn−1⋯uiyx{u}_{n}{v}_{n}{v}_{n-1}\cdots {u}_{i}yare three total proper paths connecting xxand yy. Hence, CL2n{{\rm{CL}}}_{2n}is total proper 3-connected with the above coloring.Case 2. n≡1(mod3)n\equiv 1\left({\rm{mod}}\hspace{0.33em}3). Let n=3t+1n=3t+1. Define a total coloring ccof CL2n{{\rm{CL}}}_{2n}as follows: c(u3i+1)=c(u3i+2u3i+3)=c(v3i+3)=c(v3i+1v3i+2)=1c\left({u}_{3i+1})=c\left({u}_{3i+2}{u}_{3i+3})=c\left({v}_{3i+3})=c\left({v}_{3i+1}{v}_{3i+2})=1, c(u3i+2)=c(u3i+3u3i+4)=c(v3i+2)=c(v3i+3v3i+4)=3c\left({u}_{3i+2})=c\left({u}_{3i+3}{u}_{3i+4})=c\left({v}_{3i+2})=c\left({v}_{3i+3}{v}_{3i+4})=3, and c(u3i+3)=c(u3i+1u3i+2)=c(v3i+1)=c(v3i+2v3i+3)=2c\left({u}_{3i+3})=c\left({u}_{3i+1}{u}_{3i+2})=c\left({v}_{3i+1})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{3i+2}{v}_{3i+3})\hspace{-0.08em}=\hspace{-0.08em}2, where 0≤i≤t−10\le i\le t-1; c(un)=c(v1vn)=1,c(unvn)=2c\left({u}_{n})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{1}{v}_{n})\hspace{-0.08em}=\hspace{-0.08em}1,c\left({u}_{n}{v}_{n})\hspace{-0.08em}=\hspace{-0.08em}2, c(vn)=c(u1un)=4c\left({v}_{n})\hspace{-0.08em}=\hspace{-0.08em}c\left({u}_{1}{u}_{n})\hspace{-0.08em}=\hspace{-0.08em}4, and c(ujvj)=4c\left({u}_{j}{v}_{j})=4for 2≤j≤n−12\le j\le n-1. Let xxand yybe any two distinct vertices of CL2n{{\rm{CL}}}_{2n}. By symmetry, we may assume that x=u1x={u}_{1}. If y=uiy={u}_{i}for 2≤i≤n2\le i\le n, then xu2u3⋯ui−1y,xunun−1⋯ui+1yx{u}_{2}{u}_{3}\cdots {u}_{i-1}y,x{u}_{n}{u}_{n-1}\cdots {u}_{i+1}y, and xv1v2⋯viyx{v}_{1}{v}_{2}\cdots {v}_{i}yare three total proper paths connecting xxand yy. If y=v1y={v}_{1}, then xy,xu2v2yxy,x{u}_{2}{v}_{2}yand xunvnyx{u}_{n}{v}_{n}yare three total proper paths connecting xxand yy. If y=viy={v}_{i}for 2≤i≤n2\le i\le n, then xu1u2⋯uiy,xv1v2⋯vi−1yx{u}_{1}{u}_{2}\cdots {u}_{i}y,x{v}_{1}{v}_{2}\cdots {v}_{i-1}y, and xunvnvn−1⋯uiyx{u}_{n}{v}_{n}{v}_{n-1}\cdots {u}_{i}yare three total proper paths connecting xxand yy. Hence, CL2n{{\rm{CL}}}_{2n}is total proper 3-connected with the above coloring.Case 3. n≡2(mod3)n\equiv 2\left({\rm{mod}}\hspace{0.33em}3). Let n=3t+2n=3t+2. Define a total coloring ccof CL2n{{\rm{CL}}}_{2n}as follows: c(u3i+1)=c(u3i+2u3i+3)=c(v3i+4)=c(v3i+2v3i+3)=1c\left({u}_{3i+1})=c\left({u}_{3i+2}{u}_{3i+3})=c\left({v}_{3i+4})=c\left({v}_{3i+2}{v}_{3i+3})=1, c(u3i+2)=c(u3i+3u3i+4)=c(v3i+3)=c(v3i+1v3i+2)=3c\left({u}_{3i+2})=c\left({u}_{3i+3}{u}_{3i+4})=c\left({v}_{3i+3})=c\left({v}_{3i+1}{v}_{3i+2})=3, and c(u3i+3)=c(u3i+1u3i+2)=c(v3i+2)=c(v3i+3v3i+4)=2c\left({u}_{3i+3})=c\left({u}_{3i+1}{u}_{3i+2})=c\left({v}_{3i+2})=c\left({v}_{3i+3}{v}_{3i+4})=2, where 0≤i≤t−10\le i\le t-1; c(un)=c(vn−1vn)=3,c(vn)=c(un−1un)=c(u1v1)=2c\left({u}_{n})=c\left({v}_{n-1}{v}_{n})=3,c\left({v}_{n})=c\left({u}_{n-1}{u}_{n})=c\left({u}_{1}{v}_{1})=2, c(un−1)=c(v1vn)=c(unu1)=1,c(v1)=4c\left({u}_{n-1})=c\left({v}_{1}{v}_{n})=c\left({u}_{n}{u}_{1})=1,c\left({v}_{1})=4, and c(ujvj)=4c\left({u}_{j}{v}_{j})=4for 2≤j≤n2\le j\le n. Let xxand yybe any two distinct vertices of CL2n{{\rm{CL}}}_{2n}. By symmetry, we may assume that x=u1x={u}_{1}. If y=uiy={u}_{i}for 2≤i≤n2\le i\le n, then xu2u3⋯ui−1y,xunun−1⋯ui+1yx{u}_{2}{u}_{3}\cdots {u}_{i-1}y,x{u}_{n}{u}_{n-1}\cdots {u}_{i+1}y, and xv1v2⋯viyx{v}_{1}{v}_{2}\cdots {v}_{i}yare three total proper paths connecting xxand yy. If y=v1y={v}_{1}, then xy,xu2v2yxy,x{u}_{2}{v}_{2}y, and xunvnyx{u}_{n}{v}_{n}yare three total proper paths connecting xxand yy. If y=viy={v}_{i}for 2≤i≤n2\le i\le n, then xu1u2⋯uiy,xv1v2⋯vi−1yx{u}_{1}{u}_{2}\cdots {u}_{i}y,x{v}_{1}{v}_{2}\cdots {v}_{i-1}yand xunvnvn−1⋯uiyx{u}_{n}{v}_{n}{v}_{n-1}\cdots {u}_{i}yare three total proper paths connecting xxand yy. Hence, CL2n{{\rm{CL}}}_{2n}is total proper 3-connected with the above coloring.□Next, we shall determine the total proper kk-connection numbers of the Möbius ladders.Lemma 7.2Let nnbe an integer with n≥3n\ge 3. Then tpc(M2n)=tpc2(M2n)=3,tpc3(M2n)=4{\rm{tpc}}\left({M}_{2n})={{\rm{tpc}}}_{2}\left({M}_{2n})=3,{{\rm{tpc}}}_{3}\left({M}_{2n})=4.ProofSince M2n{M}_{2n}contains a Hamiltonian path and is not complete, we have tpc(M2n)=3{\rm{tpc}}\left({M}_{2n})=3. Since tpc2(M2n)≥tpc(M2n)=3{{\rm{tpc}}}_{2}\left({M}_{2n})\ge {\rm{tpc}}\left({M}_{2n})=3, we only need to prove tpc2(M2n)≤3{{\rm{tpc}}}_{2}\left({M}_{2n})\le 3.Case 1. n≡0(mod3)n\equiv 0\left({\rm{mod}}\hspace{0.33em}3). Define a total coloring ccof M2n{M}_{2n}as follows: Let iibe an integer with 1≤i≤n−21\le i\le n-2, c(ui)=c(vn−i+1)=1,c(ui+1)=c(vn−i)=3c\left({u}_{i})=c\left({v}_{n-i+1})=1,c\left({u}_{i+1})=c\left({v}_{n-i})=3, and c(ui+2)=c(vn−i−1)=2c\left({u}_{i+2})=c\left({v}_{n-i-1})=2; c(uiui+1),c(ui),c(ui+1)∈{1,2,3}c\left({u}_{i}{u}_{i+1}),c\left({u}_{i}),c\left({u}_{i+1})\in \left\{1,2,3\right\}with c(uiui+1)≠c(ui)≠c(ui+1)c\left({u}_{i}{u}_{i+1})\ne c\left({u}_{i})\ne c\left({u}_{i+1})for 1≤i≤n−11\le i\le n-1; c(vivi+1),c(vi),c(vi+1)∈{1,2,3}c\left({v}_{i}{v}_{i+1}),c\left({v}_{i}),c\left({v}_{i+1})\in \left\{1,2,3\right\}with c(vivi+1)≠c(vi)≠c(vi+1)c\left({v}_{i}{v}_{i+1})\ne c\left({v}_{i})\ne c\left({v}_{i+1})for 1≤i≤n−11\le i\le n-1; c(u1v1)=3,c(unvn)=2c\left({u}_{1}{v}_{1})=3,c\left({u}_{n}{v}_{n})=2, and c(ujvn−j+1)=3c\left({u}_{j}{v}_{n-j+1})=3for 1≤j≤n1\le j\le n. Let xxand yybe any two distinct vertices of M2n{M}_{2n}. By symmetry, we may assume that x=u1x={u}_{1}. If y=uiy={u}_{i}for 2≤i≤n2\le i\le n, then xu2u3⋯ui−1yx{u}_{2}{u}_{3}\cdots {u}_{i-1}yand xv1v2⋯vnunun−1⋯ui+1yx{v}_{1}{v}_{2}\cdots {v}_{n}{u}_{n}{u}_{n-1}\cdots {u}_{i+1}yare two total proper paths connecting xxand yy. If y=v1y={v}_{1}, then xyxyand xu2u3yx{u}_{2}{u}_{3}yare two total proper paths connecting xxand yy. If y=viy={v}_{i}for 2≤i≤n2\le i\le n, then xu1u2⋯unvnvn−1⋯vi+1yx{u}_{1}{u}_{2}\cdots {u}_{n}{v}_{n}{v}_{n-1}\cdots {v}_{i+1}yand xv1v2⋯vi−1yx{v}_{1}{v}_{2}\cdots {v}_{i-1}yare two total proper paths connecting xxand yy. Thus, M2n{M}_{2n}is total proper 2-connected with the above coloring.Case 2. n≡1(mod3)n\equiv 1\left({\rm{mod}}\hspace{0.33em}3). Let n=3t+1n=3t+1. Define a total coloring ccof M2n{M}_{2n}as follows: Let iibe an integer with 0≤i≤t−10\le i\le t-1, c(u3i+1)=c(u3i+2u3i+3)=c(v3i+3)=c(v3i+1v3i+2)=1c\left({u}_{3i+1})=c\left({u}_{3i+2}{u}_{3i+3})=c\left({v}_{3i+3})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{3i+1}{v}_{3i+2})\hspace{-0.08em}=\hspace{-0.08em}1, c(u3i+2)=c(u3i+3u3i+4)=c(v3i+1)=c(v3i+2v3i+3)=3c\left({u}_{3i+2})\hspace{-0.08em}=\hspace{-0.08em}c\left({u}_{3i+3}{u}_{3i+4})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{3i+1})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{3i+2}{v}_{3i+3})=3, and c(u3i+3)=c(u3i+1u3i+2)=c(v3i+2)=c(v3i+3v3i+4)=2c\left({u}_{3i+3})=c\left({u}_{3i+1}{u}_{3i+2})=c\left({v}_{3i+2})=c\left({v}_{3i+3}{v}_{3i+4})=2; if j=3ij=3ifor 1≤i≤t1\le i\le t, then c(ujvn−j+1)=3c\left({u}_{j}{v}_{n-j+1})=3; if j≠3ij\ne 3ifor 1≤i≤t1\le i\le t, then c(ujvn−j+1),c(uj),c(vn−j+1)∈{1,2,3}c\left({u}_{j}{v}_{n-j+1}),c\left({u}_{j}),c\left({v}_{n-j+1})\in \left\{1,2,3\right\}with c(ujvn−j+1)≠c(uj)≠c(vn−j+1)c\left({u}_{j}{v}_{n-j+1})\ne c\left({u}_{j})\ne c\left({v}_{n-j+1}); c(un)=1,c(unvn)=c(u1v1)=2,c(vn)=3c\left({u}_{n})=1,c\left({u}_{n}{v}_{n})=c\left({u}_{1}{v}_{1})=2,c\left({v}_{n})=3. Let xxand yybe any two distinct vertices of M2n{M}_{2n}. We may assume that x=uix={u}_{i}for 1≤i≤n1\le i\le n. If y=ujy={u}_{j}for i≤j≤ni\le j\le n, then xui+1ui+2⋯uj−1yx{u}_{i+1}{u}_{i+2}\cdots {u}_{j-1}yand xui−1ui−2⋯u1vnunun−1⋯uj+1yx{u}_{i-1}{u}_{i-2}\cdots {u}_{1}{v}_{n}{u}_{n}{u}_{n-1}\cdots {u}_{j+1}yare two total proper paths connecting xxand yy. If y=vjy={v}_{j}for 1≤j≤n1\le j\le n, then xui−1ui−2⋯u1vnvn−1⋯vj+1yx{u}_{i-1}{u}_{i-2}\cdots {u}_{1}{v}_{n}{v}_{n-1}\cdots {v}_{j+1}yand xui+1⋯un−j+1yx{u}_{i+1}\cdots {u}_{n-j+1}yare two total proper paths connecting xxand yy. Thus, M2n{M}_{2n}is total proper 2-connected with the above coloring.Case 3. n≡2(mod3)n\equiv 2\left({\rm{mod}}\hspace{0.33em}3). Let n=3t+2n=3t+2. Define a total coloring ccof M2n{M}_{2n}as follows: Let iibe an integer with 0≤i≤t−10\hspace{0.08em}\le \hspace{0.08em}i\le t\hspace{0.08em}-\hspace{0.08em}1, c(u3i+1)=c(u3i+2u3i+3)=c(v3i+3)=c(v3i+1v3i+2)=c(u3i+2vn−3i−1)=1c\left({u}_{3i+1})\hspace{0.08em}=\hspace{0.08em}c\left({u}_{3i+2}{u}_{3i+3})\hspace{0.08em}=\hspace{0.08em}c\left({v}_{3i+3})=c\left({v}_{3i+1}{v}_{3i+2})=c\left({u}_{3i+2}{v}_{n-3i-1})=1, c(u3i+2)=c(u3i+3u3i+4)=c(v3i+1)=c(v3i+2v3i+3)=c(u3i+3vn−3i−2)=3c\left({u}_{3i+2})=c\left({u}_{3i+3}{u}_{3i+4})=c\left({v}_{3i+1})=c\left({v}_{3i+2}{v}_{3i+3})\hspace{-0.08em}=\hspace{-0.08em}c\left({u}_{3i+3}{v}_{n-3i-2})\hspace{-0.08em}=\hspace{-0.08em}3, and c(u3i+3)=c(u3i+1u3i+2)=c(v3i+2)=c(v3i+3v3i+4)=c(u3i+4vn−3i−3)=2c\left({u}_{3i+3})=c\left({u}_{3i+1}{u}_{3i+2})=c\left({v}_{3i+2})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{3i+3}{v}_{3i+4})\hspace{-0.08em}=\hspace{-0.08em}c\left({u}_{3i+4}{v}_{n-3i-3})=2; c(un−1)=c(unv1)=c(vn−1vn)=1c\left({u}_{n-1})=c\left({u}_{n}{v}_{1})=c\left({v}_{n-1}{v}_{n})=1, c(un)=c(vn−1)=c(u1vn)=3c\left({u}_{n})=c\left({v}_{n-1})=c\left({u}_{1}{v}_{n})=3, c(vn)=c(un−1un)=2c\left({v}_{n})=c\left({u}_{n-1}{u}_{n})=2. Let xxand yybe any two distinct vertices of M2n{M}_{2n}. We may assume that x=uix={u}_{i}for 1≤i≤n1\le i\le n. If y=ujy={u}_{j}for i≤j≤ni\le j\le n, then xui+1ui+2⋯uj−1yx{u}_{i+1}{u}_{i+2}\cdots {u}_{j-1}yand xui−1ui−2⋯u1vnunun−1⋯uj+1yx{u}_{i-1}{u}_{i-2}\cdots {u}_{1}{v}_{n}{u}_{n}{u}_{n-1}\cdots {u}_{j+1}yare two total proper paths connecting xxand yy. If y=vjy={v}_{j}for 1≤j≤n1\le j\le n, then xui−1ui−2⋯u1vnvn−1⋯vj+1yx{u}_{i-1}{u}_{i-2}\cdots {u}_{1}{v}_{n}{v}_{n-1}\cdots {v}_{j+1}yand xui+1⋯un−j+1yx{u}_{i+1}\cdots {u}_{n-j+1}yare two total proper paths connecting xxand yy. Thus, M2n{M}_{2n}is total proper 2-connected with the above coloring.To the contrary, suppose there exists a total proper 3-connected coloring ccof M2n{M}_{2n}using three colors. By considering the pair {u2,vn}\left\{{u}_{2},{v}_{n}\right\}, u2u3⋯unvn,u2u1vn{u}_{2}{u}_{3}\cdots {u}_{n}{v}_{n},{u}_{2}{u}_{1}{v}_{n}, and u2vn−1vn{u}_{2}{v}_{n-1}{v}_{n}must be three total proper paths connecting u2{u}_{2}and vn{v}_{n}. Then c(u2vn−1)≠c(vn−1vn)≠c(vn−1)c\left({u}_{2}{v}_{n-1})\ne c\left({v}_{n-1}{v}_{n})\ne c\left({v}_{n-1}). By considering the pair {u2,vn−2}\left\{{u}_{2},{v}_{n-2}\right\}, u2u1vnvn−1vn−2,u2u3vn−2{u}_{2}{u}_{1}{v}_{n}{v}_{n-1}{v}_{n-2},{u}_{2}{u}_{3}{v}_{n-2}, and u2vn−1vn{u}_{2}{v}_{n-1}{v}_{n}must be three total proper paths connecting u2{u}_{2}and vn−2{v}_{n-2}. Thus, c(u2vn−1)≠c(vn−1vn−2)≠c(vn−1)c\left({u}_{2}{v}_{n-1})\ne c\left({v}_{n-1}{v}_{n-2})\ne c\left({v}_{n-1}), and hence c(vn−1vn−2)=c(vn−1vn)c\left({v}_{n-1}{v}_{n-2})=c\left({v}_{n-1}{v}_{n}). But then, there is no set of three disjoint total proper paths connecting vn−2{v}_{n-2}and vn{v}_{n}, a contradiction. Thus, tpc3(M2n)≥4{{\rm{tpc}}}_{3}\left({M}_{2n})\ge 4. Now we only need to prove tpc3(M2n)≤4{{\rm{tpc}}}_{3}\left({M}_{2n})\le 4.Case 1. n≡0(mod2)n\equiv 0\left({\rm{mod}}\hspace{0.33em}2). Let n=2tn=2t. Assign a total coloring ccof M2n{M}_{2n}as follows: c(u2i+1)=c(vn−2i)=c(u2i+2vn−2i−1)=1c\left({u}_{2i+1})=c\left({v}_{n-2i})=c\left({u}_{2i+2}{v}_{n-2i-1})\hspace{-0.1em}=\hspace{-0.1em}1, c(u2i+2)=c(un−2i−1)=c(u2i+1vn−2i)=3c\left({u}_{2i+2})=c\left({u}_{n-2i-1})=c\left({u}_{2i+1}{v}_{n-2i})=3, c(vn−2ivn−2i−1)=c(u2i+1u2i+2)=2c\left({v}_{n-2i}{v}_{n-2i-1})\hspace{-0.1em}=\hspace{-0.1em}c\left({u}_{2i+1}{u}_{2i+2})=2, where 0≤i≤t−10\le i\le t-1; c(u2i+2u2i+3)=c(vn−2i−1vn−2i−2)=4c\left({u}_{2i+2}{u}_{2i+3})=c\left({v}_{n-2i-1}{v}_{n-2i-2})=4for 0≤i≤t−20\le i\le t-2; c(unvn)=c(u1v1)=4c\left({u}_{n}{v}_{n})=c\left({u}_{1}{v}_{1})=4. Let xxand yybe any two distinct vertices of M2n{M}_{2n}. By symmetry, we may assume that x=u1x={u}_{1}. If y=uiy={u}_{i}for 2≤i≤n2\le i\le n, then xu2⋯ui−1y,xvnunun−1⋯ui+1yx{u}_{2}\cdots {u}_{i-1}y,x{v}_{n}{u}_{n}{u}_{n-1}\cdots {u}_{i+1}y, and xv1v2⋯vn−i+1uix{v}_{1}{v}_{2}\cdots {v}_{n-i+1}{u}_{i}are three total proper paths connecting xxand yy. If y=v1y={v}_{1}, then xy,xu2u3⋯unyxy,x{u}_{2}{u}_{3}\cdots {u}_{n}y, and xvnvn−1⋯v2yx{v}_{n}{v}_{n-1}\cdots {v}_{2}yare three total proper paths connecting xxand yy. If y=viy={v}_{i}for 2≤i≤n2\le i\le n, then xv1v2⋯vi−1y,xvnvn−1⋯vi+1yx{v}_{1}{v}_{2}\cdots {v}_{i-1}y,x{v}_{n}{v}_{n-1}\cdots {v}_{i+1}yand xu2u3⋯un−i+1yx{u}_{2}{u}_{3}\cdots {u}_{n-i+1}yare three total proper paths connecting xxand yy. Hence, M2n{M}_{2n}is total proper 3-connected with the above coloring.Case 2. n≡1(mod2)n\equiv 1\left({\rm{mod}}\hspace{0.33em}2). For n=3n=3, we assign a total coloring ccto M6{M}_{6}as follows: c(u1)=c(v3)=c(u2u3)=c(v1v2)=1c\left({u}_{1})=c\left({v}_{3})=c\left({u}_{2}{u}_{3})=c\left({v}_{1}{v}_{2})\hspace{-0.1em}=\hspace{-0.1em}1, c(u2)=c(v2)=c(u3v3)=c(u1v1)=2c\left({u}_{2})=c\left({v}_{2})=c\left({u}_{3}{v}_{3})=c\left({u}_{1}{v}_{1})=2, c(u3)=c(v1)=c(u1u2)=c(v2v3)=3c\left({u}_{3})=c\left({v}_{1})=c\left({u}_{1}{u}_{2})\hspace{-0.1em}=\hspace{-0.1em}c\left({v}_{2}{v}_{3})=3, c(uivn−i+1)=4c\left({u}_{i}{v}_{n-i+1})=4for 1≤i≤31\le i\le 3. We can verify that M6{M}_{6}is total proper 3-connected with the above total coloring, and so tpc3(M6)=4{{\rm{tpc}}}_{3}\left({M}_{6})=4.For n=5n=5, define a total coloring ccof M10{M}_{10}as follows: c(u1)=c(v5)=1c\left({u}_{1})=c\left({v}_{5})=1, c(u2)=c(v4)=2c\left({u}_{2})=c\left({v}_{4})=2, c(u3)=c(v3)=c(u5)=c(v1)=3c\left({u}_{3})=c\left({v}_{3})=c\left({u}_{5})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{1})\hspace{-0.08em}=\hspace{-0.08em}3, c(u4)=c(v2)=4c\left({u}_{4})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{2})\hspace{-0.08em}=\hspace{-0.08em}4, c(u2u3)=c(u4u5)=c(v1v2)=c(v3v4)=1c\left({u}_{2}{u}_{3})\hspace{-0.1em}=\hspace{-0.1em}c\left({u}_{4}{u}_{5})\hspace{-0.1em}=\hspace{-0.1em}c\left({v}_{1}{v}_{2})\hspace{-0.1em}=\hspace{-0.1em}c\left({v}_{3}{v}_{4})\hspace{-0.1em}=\hspace{-0.1em}1, c(u3u4)=c(u5v5)=c(u1v1)=c(v2v3)=2c\left({u}_{3}{u}_{4})\hspace{-0.1em}=\hspace{-0.1em}c\left({u}_{5}{v}_{5})\hspace{-0.1em}=\hspace{-0.1em}c\left({u}_{1}{v}_{1})\hspace{-0.1em}=\hspace{-0.1em}c\left({v}_{2}{v}_{3})\hspace{-0.1em}=2, c(u4v2)=c(u1u2)=c(v4v5)=3c\left({u}_{4}{v}_{2})=c\left({u}_{1}{u}_{2})=c\left({v}_{4}{v}_{5})=3, c(uivn−i+1)=4c\left({u}_{i}{v}_{n-i+1})=4for i=1,2,3,5i=1,2,3,5. We can check that M10{M}_{10}is total proper 3-connected with the above total coloring, and so tpc3(M10)=4{{\rm{tpc}}}_{3}\left({M}_{10})=4.Subcase 2.1. Let n≡1(mod4)n\equiv 1\left({\rm{mod}}\hspace{0.33em}4)for n≥7n\ge 7. Let n=2t+1n=2t+1. Assign a total coloring ccto M2n{M}_{2n}as follows: c(u4i+1)=c(vn−4i)=1c\left({u}_{4i+1})=c\left({v}_{n-4i})=1, c(u4i+2)=c(vn−4i−1)=2c\left({u}_{4i+2})=c\left({v}_{n-4i-1})=2, c(u4i+3)=c(vn−4i−2)=3c\left({u}_{4i+3})=c\left({v}_{n-4i-2})=3, and c(u4i+4)=c(vn−4i−3)=4c\left({u}_{4i+4})=c\left({v}_{n-4i-3})=4, where 0≤i≤t−220\le i\le \frac{t-2}{2}; c(u4iu4i+1)=c(vn−4i+1vn−4i)=c(u4i+2vn−4i−1)=3c\left({u}_{4i}{u}_{4i+1})=c\left({v}_{n-4i+1}{v}_{n-4i})=c\left({u}_{4i+2}{v}_{n-4i-1})\hspace{-0.08em}=\hspace{-0.08em}3, c(u4i+1u4i+2)=c(vn−4ivn−4i−1)=c(u4i+3vn−4i−2)=4c\left({u}_{4i+1}{u}_{4i+2})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{n-4i}{v}_{n-4i-1})\hspace{-0.08em}=\hspace{-0.08em}c\left({u}_{4i+3}{v}_{n-4i-2})\hspace{-0.08em}=\hspace{-0.08em}4, c(u4i+2u4i+3)=c(vn−4i−1vn−4i−2)=c(u4ivn−4i+1)=1c\left({u}_{4i+2}{u}_{4i+3})\hspace{0.1em}=\hspace{0.1em}c\left({v}_{n-4i-1}{v}_{n-4i-2})=c\left({u}_{4i}{v}_{n-4i+1})\hspace{0.1em}=\hspace{0.1em}1, and c(vn−4i−2vn−4i−3)=c(u4i+3u4i+4)=c(u4i+1vn−4i)=2c\left({v}_{n-4i-2}{v}_{n-4i-3})\hspace{0.1em}=\hspace{0.1em}c\left({u}_{4i+3}{u}_{4i+4})\hspace{0.1em}=\hspace{0.1em}c\left({u}_{4i+1}{v}_{n-4i})\hspace{0.1em}=\hspace{0.1em}2, where 1≤i≤t−221\hspace{-0.08em}\le \hspace{-0.08em}i\hspace{-0.08em}\le \hspace{-0.08em}\frac{t-2}{2}; c(un)=c(v1)=c(u1u2)=c(vnvn−1)=3c\left({u}_{n})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{1})\hspace{-0.08em}=\hspace{-0.08em}c\left({u}_{1}{u}_{2})\hspace{-0.08em}=\hspace{-0.1em}c\left({v}_{n}{v}_{n-1})=3, c(u2u3)=c(vn−1vn−2)=c(v1v2)=c(un−1un)=c(un−1v2)=1c\left({u}_{2}{u}_{3})=c\left({v}_{n-1}{v}_{n-2})=c\left({v}_{1}{v}_{2})=c\left({u}_{n-1}{u}_{n})=c\left({u}_{n-1}{v}_{2})=1, c(u3u4)=c(vn−2vn−3)=c(u1v1)=c(unvn)=2c\left({u}_{3}{u}_{4})=c\left({v}_{n-2}{v}_{n-3})=c\left({u}_{1}{v}_{1})=c\left({u}_{n}{v}_{n})=2, and c(u1vn)=c(u2vn−1)=c(u3vn−2)=4c\left({u}_{1}{v}_{n})=c\left({u}_{2}{v}_{n-1})=c\left({u}_{3}{v}_{n-2})=4. Let xxand yybe any two distinct vertices of M2n{M}_{2n}. By symmetry, we may assume that x=u1x={u}_{1}. If y=uiy={u}_{i}for 2≤i≤n2\le i\le n, then xu2u3⋯ui−1y,xvnunun−1⋯ui+1yx{u}_{2}{u}_{3}\cdots {u}_{i-1}y,x{v}_{n}{u}_{n}{u}_{n-1}\cdots {u}_{i+1}y, and xv1v2⋯vn−i+1uix{v}_{1}{v}_{2}\cdots {v}_{n-i+1}{u}_{i}are three total proper paths connecting xxand yy. If y=v1y={v}_{1}, then xy,xu2u3⋯unyxy,x{u}_{2}{u}_{3}\cdots {u}_{n}y, and xvnvn−1⋯v2yx{v}_{n}{v}_{n-1}\cdots {v}_{2}yare three total proper paths connecting xxand yy. If y=viy={v}_{i}for 2≤i≤n2\le i\le n, then xv1v2⋯vi−1y,xvnvn−1⋯vi+1yx{v}_{1}{v}_{2}\cdots {v}_{i-1}y,x{v}_{n}{v}_{n-1}\cdots {v}_{i+1}yand xu2u3⋯un−i+1yx{u}_{2}{u}_{3}\cdots {u}_{n-i+1}yare three total proper paths connecting xxand yy. Hence, M2n{M}_{2n}is total proper 3-connected with the above coloring.Subcase 2.2. Let n≡3(mod4)n\equiv 3\left({\rm{mod}}\hspace{0.33em}4)for n≥7n\ge 7. Let n=2t+1n=2t+1. Assign a total coloring ccto M2n{M}_{2n}as follows: Let iibe an integer with 0≤i≤t−320\le i\le \frac{t-3}{2}, c(u4i+1)=c(vn−4i)=1c\left({u}_{4i+1})=c\left({v}_{n-4i})=1, c(u4i+2)=c(vn−4i−1)=2c\left({u}_{4i+2})=c\left({v}_{n-4i-1})=2, c(u4i+3)=c(vn−4i−2)=3c\left({u}_{4i+3})=c\left({v}_{n-4i-2})=3, and c(u4i+4)=c(vn−4i−3)=4c\left({u}_{4i+4})=c\left({v}_{n-4i-3})=4. Let iibe an integer with 1≤i≤t−121\le i\le \frac{t-1}{2}, c(u4iu4i+1)=c(vn−4i−1vn−4i)=c(u4i+2vn−4i−1)=3c\left({u}_{4i}{u}_{4i+1})=c\left({v}_{n-4i-1}{v}_{n-4i})=c\left({u}_{4i+2}{v}_{n-4i-1})=3, c(u4i+1u4i+2)=c(vn−4ivn−4i−1)=c(u4i+3vn−4i−2)=4c\left({u}_{4i+1}{u}_{4i+2})\hspace{0.1em}=\hspace{0.1em}c\left({v}_{n-4i}{v}_{n-4i-1})\hspace{0.1em}=\hspace{0.1em}c\left({u}_{4i+3}{v}_{n-4i-2})\hspace{0.1em}=\hspace{0.1em}4, c(u4i+2u4i+3)=c(u4ivn−4i+1)=c(vn−4i−1vn−4i−2)=1c\left({u}_{4i+2}{u}_{4i+3})\hspace{0.1em}=\hspace{0.1em}c\left({u}_{4i}{v}_{n-4i+1})\hspace{0.12em}=\hspace{0.12em}c\left({v}_{n-4i-1}{v}_{n-4i-2})\hspace{0.12em}=\hspace{0.12em}1, and c(u4i+1vn−4i)=c(u4i−1u4i)=c(vn−4i+2vn−4i+1)=2c\left({u}_{4i+1}{v}_{n-4i})\hspace{-0.1em}=\hspace{-0.1em}c\left({u}_{4i-1}{u}_{4i})\hspace{-0.1em}=\hspace{-0.1em}c\left({v}_{n-4i+2}{v}_{n-4i+1})\hspace{-0.1em}=\hspace{-0.1em}2; c(un)=c(v1)=c(u1u2)=c(vnvn−1)=3c\left({u}_{n})\hspace{-0.1em}=\hspace{-0.1em}c\left({v}_{1})\hspace{-0.1em}=\hspace{-0.1em}c\left({u}_{1}{u}_{2})\hspace{-0.1em}=\hspace{-0.1em}c\left({v}_{n}{v}_{n-1})\hspace{-0.1em}=\hspace{-0.1em}3, c(un−1)=c(v2)=c(u1v1)=c(unvn)=2c\left({u}_{n-1})\hspace{-0.08em}=\hspace{-0.08em}c\left({v}_{2})\hspace{-0.08em}=c\left({u}_{1}{v}_{1})\hspace{-0.08em}=c\left({u}_{n}{v}_{n})=2, and c(un−2)=c(v3)=c(u2u3)=c(vn−1vn−2)=1c\left({u}_{n-2})=c\left({v}_{3})=c\left({u}_{2}{u}_{3})=c\left({v}_{n-1}{v}_{n-2})=1. By the similar proof of the above subcase, we can verify M2n{M}_{2n}is total proper 3-connected with the above coloring.□8Proof of Theorem 1.7Note that K3□K2=CL6,K3,3=M6,Q3=CL8{K}_{3}\hspace{0.33em}\square \hspace{0.33em}{K}_{2}={{\rm{CL}}}_{6},{K}_{3,3}={M}_{6},{Q}_{3}={{\rm{CL}}}_{8}and M8{M}_{8}. By means of Theorem 1.6, we can obtain their total proper kk-connection numbers. Now, we only need to consider K4,F1,F2,F3{K}_{4},{F}_{1},{F}_{2},{F}_{3}.Lemma 8.1tpc(K4)=1,tpc2(K4)=3,tpc3(K4)=4{\rm{tpc}}\left({K}_{4})=1,{{\rm{tpc}}}_{2}\left({K}_{4})=3,{{\rm{tpc}}}_{3}\left({K}_{4})=4.ProofBy [26], we know that tpc(K4)=1{\rm{tpc}}\left({K}_{4})=1. Suppose tpc2(K4)=2{{\rm{tpc}}}_{2}\left({K}_{4})=2, then there is no set of two disjoint total proper paths connecting u1{u}_{1}and u2{u}_{2}, a contradiction. Thus, tpc2(K4)≥3{{\rm{tpc}}}_{2}\left({K}_{4})\ge 3. Let V(K4)={u1,u2,u3,u4}V\left({K}_{4})=\left\{{u}_{1},{u}_{2},{u}_{3},{u}_{4}\right\}, we assign a total coloring ccto K4{K}_{4}as follows: c(u1)=c(u2u3)=c(u3u4)=c(u2u4)=1c\left({u}_{1})=c\left({u}_{2}{u}_{3})=c\left({u}_{3}{u}_{4})=c\left({u}_{2}{u}_{4})=1, c(u2)=c(u4)=c(u1u3)=2c\left({u}_{2})=c\left({u}_{4})=c\left({u}_{1}{u}_{3})=2, c(u3)=c(u1u2)=c(u1u4)=3c\left({u}_{3})=c\left({u}_{1}{u}_{2})=c\left({u}_{1}{u}_{4})=3. We can verify that the K4{K}_{4}is total proper 2-connected, so tpc2(K4)=3{{\rm{tpc}}}_{2}\left({K}_{4})=3.Now, we suppose there exists a total proper 3-connected coloring ccof K4{K}_{4}using three colors. Considering u1{u}_{1}and u2{u}_{2}, u1u2{u}_{1}{u}_{2}, u1u4u2{u}_{1}{u}_{4}{u}_{2}, and u1u3u2{u}_{1}{u}_{3}{u}_{2}must be the three total proper paths connecting u1{u}_{1}and u2{u}_{2}. Then c(u1u4)≠c(u2u4)≠c(u4)c\left({u}_{1}{u}_{4})\ne c\left({u}_{2}{u}_{4})\ne c\left({u}_{4}). Considering u1{u}_{1}and u3{u}_{3}, u1u3{u}_{1}{u}_{3}, u1u2u3{u}_{1}{u}_{2}{u}_{3}, and u1u4u3{u}_{1}{u}_{4}{u}_{3}must be the three total proper paths connecting u1{u}_{1}and u3{u}_{3}. Thus, c(u1u4)≠c(u3u4)≠c(u4)c\left({u}_{1}{u}_{4})\ne c\left({u}_{3}{u}_{4})\ne c\left({u}_{4}), and so c(u2u4)=c(u4u3)c\left({u}_{2}{u}_{4})=c\left({u}_{4}{u}_{3}). But then, there is no set of three disjoint total proper paths connecting u2{u}_{2}and u3{u}_{3}, a contradiction. Hence, tpc3(K4)≥4{{\rm{tpc}}}_{3}\left({K}_{4})\ge 4. Define a total coloring ccof K4{K}_{4}as follows: c(u1)=c(u3)=1c\left({u}_{1})=c\left({u}_{3})=1, c(u2)=c(u4)=2c\left({u}_{2})=c\left({u}_{4})=2, c(u1u2)=c(u3u4)=4c\left({u}_{1}{u}_{2})=c\left({u}_{3}{u}_{4})=4, c(u1u4)=c(u2u3)=3c\left({u}_{1}{u}_{4})=c\left({u}_{2}{u}_{3})=3, c(u1u3)=2c\left({u}_{1}{u}_{3})=2, c(u2u4)=1c\left({u}_{2}{u}_{4})=1. We can easily check that K4{K}_{4}is total proper 3-connected, and so tpc3(K4)=4{{\rm{tpc}}}_{3}\left({K}_{4})=4.□Lemma 8.2tpc(F1)=tpc2(F1)=3,tpc3(F1)=4{\rm{tpc}}\left({F}_{1})={{\rm{tpc}}}_{2}\left({F}_{1})=3,{{\rm{tpc}}}_{3}\left({F}_{1})=4.Figure 4The total proper k-connected coloring of F1, F2 and F3.ProofSince F1{F}_{1}has a Hamiltonian path that is not complete, we know that tpc(F1)=3{\rm{tpc}}\left({F}_{1})=3. It is easy to verify that F1{F}_{1}is total proper 2-connected depicted in Figure 4(a), and so tpc2(F1)=3{{\rm{tpc}}}_{2}\left({F}_{1})=3. Now, we suppose there exists a total proper 3-connected coloring ccof F1{F}_{1}using three colors. By considering the pair {u2,u8}\left\{{u}_{2},{u}_{8}\right\}, u2u8,u2u1u8{u}_{2}{u}_{8},{u}_{2}{u}_{1}{u}_{8}, and u2u3⋯u8{u}_{2}{u}_{3}\cdots {u}_{8}must be the three total proper paths connecting u2{u}_{2}and u8{u}_{8}. Then c(u1u2)≠c(u1u8)≠c(u1)c\left({u}_{1}{u}_{2})\ne c\left({u}_{1}{u}_{8})\ne c\left({u}_{1}). By considering the pair {u5,u8}\left\{{u}_{5},{u}_{8}\right\}, u5u6u7u8,u5u1u8{u}_{5}{u}_{6}{u}_{7}{u}_{8},{u}_{5}{u}_{1}{u}_{8}, and u5u4u3u2u8{u}_{5}{u}_{4}{u}_{3}{u}_{2}{u}_{8}must be the three total proper paths connecting u5{u}_{5}and u8{u}_{8}. Then c(u1u5)≠c(u1u8)≠c(u1)c\left({u}_{1}{u}_{5})\ne c\left({u}_{1}{u}_{8})\ne c\left({u}_{1}), and hence c(u1u2)=c(u1u5)c\left({u}_{1}{u}_{2})=c\left({u}_{1}{u}_{5}). But then, there is no set of three disjoint total proper paths connecting u2{u}_{2}and u5{u}_{5}, a contradiction. Hence, tpc3(F1)≥4{{\rm{tpc}}}_{3}\left({F}_{1})\ge 4. By Figure 4(b), we know that F1{F}_{1}is total proper 3-connected, and so tpc3(F1)=4{{\rm{tpc}}}_{3}\left({F}_{1})=4.□Lemma 8.3tpc(F2)=tpc2(F2)=3{\rm{tpc}}\left({F}_{2})={{\rm{tpc}}}_{2}\left({F}_{2})=3.ProofSince F1{F}_{1}has a Hamiltonian path that is not complete, we know that tpc(F2)=3{\rm{tpc}}\left({F}_{2})=3. We can check that the coloring shown in Figure 4(c) is total proper 2-connected. Thus, tpc2(F2)=3{{\rm{tpc}}}_{2}\left({F}_{2})=3.□Lemma 8.4tpc(F3)=tpc2(F3)=3,tpc3(F3)=4{\rm{tpc}}\left({F}_{3})={{\rm{tpc}}}_{2}\left({F}_{3})=3,{{\rm{tpc}}}_{3}\left({F}_{3})=4.ProofSince F3{F}_{3}has a Hamiltonian path that is not complete, we know that tpc(F3)=3{\rm{tpc}}\left({F}_{3})=3. It is easy to check that the coloring shown in Figure 4(d) is total proper 2-connected using three colors. Thus, tpc2(F3)=3{{\rm{tpc}}}_{2}\left({F}_{3})=3. Now, we suppose there exists a total proper 3-connected coloring ccof F3{F}_{3}using three colors. Considering u2{u}_{2}and u8{u}_{8}, u2u8,u2u1u8{u}_{2}{u}_{8},{u}_{2}{u}_{1}{u}_{8}and u2u3⋯u8{u}_{2}{u}_{3}\cdots {u}_{8}must be the three total proper paths connecting u2{u}_{2}and u8{u}_{8}. Then c(u1u2)≠c(u1u8)≠c(u1)c\left({u}_{1}{u}_{2})\ne c\left({u}_{1}{u}_{8})\ne c\left({u}_{1}). Considering u5{u}_{5}and u8{u}_{8}, u5u6u7u8,u5u1u8{u}_{5}{u}_{6}{u}_{7}{u}_{8},{u}_{5}{u}_{1}{u}_{8}, and u5u4u3u2u8{u}_{5}{u}_{4}{u}_{3}{u}_{2}{u}_{8}must be the three total proper paths connecting u5{u}_{5}and u8{u}_{8}. Thus, c(u1u5)≠c(u1u8)≠c(u1)c\left({u}_{1}{u}_{5})\ne c\left({u}_{1}{u}_{8})\ne c\left({u}_{1}), and so c(u1u2)=c(u1u5)c\left({u}_{1}{u}_{2})=c\left({u}_{1}{u}_{5}). But then, there is no set of three disjoint total proper paths connecting u2{u}_{2}and u5{u}_{5}, a contradiction. Hence, tpc3(F3)≥4{{\rm{tpc}}}_{3}\left({F}_{3})\ge 4. By Figure 4(e), we know that F3{F}_{3}is total proper 3-connected, and so tpc3(F3)=4{{\rm{tpc}}}_{3}\left({F}_{3})=4.□

Journal

Open Mathematicsde Gruyter

Published: Jan 1, 2022

Keywords: total coloring; total proper path; total proper k -connected; total proper k -connection number; complement graph; clique number; 05C15; 05C35; 05C40

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