Revisiting the Hamiltonian Theme in the Square of a Block: The Case of DT-Graphs

TL;DR

This paper explores Hamiltonian paths in the square of 2-connected graphs, extending known properties from 2-connected graphs to DT-graphs and general graphs, and clarifies limitations for larger k values.

Contribution

It proves that all 2-connected DT-graphs have the F_4 property and generalizes this to all 2-connected graphs, resolving an open problem.

Findings

Every 2-connected DT-graph has the F_4 property.

The F_k property does not hold for all 2-connected graphs when k ≥ 5.

The result confirms the existence of 2-connected graphs lacking the F_k property for large k.

Abstract

The square of a graph G, denoted G^2, is the graph obtained from G by joining by an edge any two nonadjacent vertices which have a common neighbor. A graph G is said to have the F_k property if for any set of k distinct vertices {x_1, x_2, ..., x_k} in G, there is a hamiltonian path from x_1 to x_2 in G^2 containing k-2 distinct edges of G of the form x_iz_i, i = 3, ..., k. It was proved many years ago that every 2-connected graph has the F_3 property. In the first part of this work, we extend this result by proving that every 2-connected DT-graph has the F_4 property (Theorem 2) and will show in the second part that this generalization holds for arbitrary 2-connected graphs, and that there exist 2-connected graphs which do not have the F_k property for any natural number k >= 5. Altogether, this answers a problem raised before in the affirmative.

Equations

$$(H^{\prime }-c^{\ast })\cap \{v,w\}=\varnothing$$

$$G^{\prime }=G_1^{\prime }\dot{\cup }{G}_2^{\prime }$$

$$G_i^{\prime }\cap \{v,w\}\ne \varnothing ,i=1,2.$$

$$K^{+}=y{x}_1{z}_1\dots u_4{x}_4{v}_4\dots u_3{x}_3{v}_3\dots z_2{x}_{2}y.$$

$$(P(x_1,x_2)-\{x_1{w}_1,x_2{w}_2\})\cup \{x_1{x}_4,x_2{x}_3,x_4{w}_1,x_3{w}_2\}$$

$$(C^{-}-\{x_1{w}_1,x_1{t}_1,x_2{w}_2\})\cup \{w_1{t}_1,x_1{x}_4{x}_3{w}_2\}\text{/}$$

$$P(x_1,c)\cup (P(c,x_2)-u{u}^{\prime })\cup \{u^{\prime }{x}_4,x_{4}u\}\text{/}$$

$$(P_{1,2}\cup C_u-\{w_2{x}_2,z_2{x}_2,u{x}_3\})\cup \{w_2{z}_2,u{x}_4,x_4{x}_3\}$$

$$K^{+}=y{x}_1{z}_1\cdots u_4{x}_4{v}_4\cdots z_2{x}_{2}y,$$

$$P=P_1\cup \{w_3^{+}{x}_3,x_3{x}_3^{+}\}-\{x_3^{+}{w}_3^{+}\}$$

$$HP=(H{P}^{-}-\{x_1{w}_1,x_2{u}_2\})\cup \{x_1{x}_4{w}_1,x_2{x}_3{u}_2\}$$

$$HP=(H{P}^{\prime }-x_1{w}_1)\cup \{x_1{x}_3{w}_1\}$$

$$HP=(H{P}^{\prime }-x_2{u}_2)\cup \{x_2{x}_3{u}_2\}.$$

$$K^{+}=y{x}_1\dots z_3{x}_3{w}_3\dots z_4{x}_4{x}_{2}y.$$

$$\tilde{G}:=\hat{G}\cup \{u_2{x}_2\}=G-(\{x_2{x}_4\}\cup G_3).$$

$$P(x_1,x_2)=x_1{x}_3{w}_1(P_{1,2}-\{x_1,x_2\})u_2{x}_4{x}_2$$

$$P(x_1,x_2)=x_1{x}_3{x}_4(P_{4,1}-x_1)s_1{w}_1(P_{1,2}-x_1)$$

$$K^{+}=y{x}_1{u}_1\cdots x_3\cdots u_2{x}_{2}y.\text{/}$$

$$K^{\ast }=(K^{\prime }-w_2{x}_2)\cup P\cup \{r_2{x}_2\}$$

$$K_y=y{x}_1{w}_1^{\prime }\cdots x_3\cdots w_2^{\prime }{x}_{2}y.$$

$$K_y=y{x}_1{w}_1^{\prime }\cdots w_2^{\prime }{x}_{2}y.$$

$$K^{+}=y{x}_1{u}_1\cdots u_3{x}_3{v}_3\cdots u_2{x}_{2}y$$

$$P(x_1,x_2)=x_1\cdots st\cdots x_2$$

$$x_1\cdots s{x}_{4}t\cdots x_2$$

$$(P(x_1,x_2)-u_2{x}_2)u_2{v}_3{x}_3{x}_2$$

$$K^{+}=y{x}_1{u}_1\cdots z_3{x}_3\cdots x_4{z}_4\cdots u_2{x}_{2}y.$$

$$H=(G^{+}-\{y,x_1,x_3\})\cup \{x,x{v}_1,x{x}_2\}.$$

Full text

Keywords: hamiltonian cycles; hamiltonian paths; square of a block

2010 Mathematics Subject Classification: 05C38,05C45

1 Introduction and History

All concepts not defined in this paper can be found in the book by Bondy and Murty, [1], or in the other references. However, we prefer definitions as given in Fleischner’s papers if they differ from the ones given in [1]. In particular, we define a graph to be eulerian if its vertices have even degree only; that is, it is not necessarily connected. This is in line with D. König’s original definition of an Eulerian graph, [12], and this is how eulerian graphs have been defined in Fleischner’s papers quoted below (many authors call such graphs even graphs, whereas they consider a graph to be eulerian if it is a connected even graph). In any case, we consider finite loopless graphs only, but allow for multiple edges which may arise in certain constructions.

The study of hamiltonian cycles and hamiltonian paths in powers of graphs goes back to the late 1950s/early 1960s and was initiated by M. Sekanina who studied certain orderings of the vertices of a given graph. In fact, he showed in [17] that the vertices of a connected graph $G\/$ of order $n\/$ can be written as a sequence $a=a_{1},a_{2},\ldots,a_{n}=b\/$ for any given $a,b\in V(G)\/$, such that the distance $d_{G}(a_{i},a_{i+1})\leq 3\/$, $i=1,\ldots,n-1\/$. This led to the general definition of the $k\/$-th power of a graph $G\/$, denoted by $G^{k}\/$, as the graph with $V(G^{k})=V(G)\/$ and $xy\in E(G^{k})\/$ if and only if $d_{G}(x,y)\leq k\/$. Thus Sekanina’s result says that $G^{3}\/$ is hamiltonian connected for every connected graph $G\/$.

Unfortunately, this result cannot be generalized to hold for $G^{2}\/$, the square of an arbitrary connected graph $G\/$ (the square of the subdivision graph of $K_{1,3}\/$ is not hamiltonian). Thus Sekanina asked in $1963\/$ at the Graph Theory Symposium in Smolenice, which graphs have a hamiltonian square, [18]. In $1964\/$, Neuman, [15], showed, however, that a tree has a hamiltonian square if and only if it is a caterpillar. On the other hand, it wasn’t until $1978\/$ when it was shown in ([19]), that Sekanina’s question was too general, for it was tantamount to asking which graphs are hamiltonian (that is, an $NP\/$-complete problem).

However, in $1966\/$ at the Graph Theory Colloquium in Tihany, Hungary, C. St. J. A. Nash-Williams asked whether it is true that $G^{2}\/$ is hamiltonian if $G\/$ is $2\/$-connected, [14], and noted that L.W. Beineke and M.D. Plummer had thought of this problem independently as well.

By the end of $1970\/$, the third author of this paper answered Nash-Williams’ question in the affirmative; the corresponding papers [5, 6] were published in $1974\/$. In the same year, it was shown that this result implied that $G^{2}\/$ is hamiltonian connected for a $2\/$-connected graph $G\/$, [2].

Further related research was triggered by Bondy’s question (asked in $1971\/$ at the Graph Theory Conference in Baton Rouge), whether hamiltonicity in $G^{2}\/$ implies that $G^{2}\/$ is vertex pancyclic (i.e., for every $v\in V(G)\/$ there are cycles of any length from $3\/$ through $|V(G)|\/$). In fact, Hobbs showed in $1976\/$, [11], that Bondy’s question has an affirmative answer for the square of $2\/$-connected graphs and connected bridgeless $DT\/$-graphs (the latter type of graphs in which every edge is incident to a vertex of degree two, was essential for answering Nash- Williams’ question – and it is essential for the main proofs of the current paper as well). The same issue of JCT B contains, however, a paper by Faudree and Schelp, [9], in which they proved for the same classes of graphs, that since $G^{2}\/$ is hamiltonian connected, there are paths joining $v\/$ and $w\/$ of arbitrary length from $d_{G^{2}}(v,w)\/$ through $|V(G)|-1\/$ for any $v,w\in V(G)\/$ (that is, $G^{2}\/$ is panconnected). They asked, however, whether this is a general phenomenon in the square of graphs (i.e., hamiltonian connectedness in $G^{2}\/$ implies panconnectedness in $G^{2}\/$). Bondy’s question and the question by Faudree and Schelp were answered in full in [7].

Already in $1973\/$ (and published in $1975\/$) the most general block-cutvertex structure was determined such that every graph within this structure has a hamiltonian total graph, [8].

In the second part of the current work we establish in [3] the strongest possible results in some sense (${\cal F}_{k}$-property), for the square of a block to be hamiltonian connected. As for hamiltonicity in the square of a block, the strongest possible result is cited Theorem E ([7, Theorem 3]). Altogether, these results will enable us to establish (in joint work with others) the most general block-cutvertex structure such that if $G\/$ satisfies this structure then $G^{2}\/$ is hamiltonian connected or at least hamiltonian. That is, what has been achieved for total graphs, [8], will be achieved for general graphs correspondingly. Here, but also in the papers [5, 6, 7, 8] the concept of $EPS\/$-graphs plays a central role; and some of the theorems in the subsequent paper [3] require intricate proofs involving explicitly or implicitly $EPS\/$-graphs.

We are fully aware that there are shorter proofs on the existence of hamiltonian cycles in the square of a block; one has been found by Říha, [16]; and more recently, a still shorter proof was found by Georgakopoulos, [10]. Moreover, a short proof of Theorem E (cited below) has been found by Müttel and Rautenbach, [13]. Unfortunately, their methods of proof do not seem to yield the special results which we can achieve with the help of $EPS\/$-graphs. This is not entirely surprising: [8, Theorem 1] states that for a graph $G\/$, the total graph $T(G)\/$ is hamiltonian if and only if $G\/$ has an $EPS\/$-graph (note that the total graph of $G\/$ is the square of the subdivision graph of $G\/$).

2 Preliminary Discussion

By a $uv\/$-path we mean a path from $u\/$ to $v\/$. If a $uv\/$-path is hamiltonian, we call it a $uv\/$-hamiltonian path.

Definition 1

Let $\/G\/$ be a graph and let $\/A=\{x_{1},x_{2},\ldots,x_{k}\}\/$ be a set of $\/k\geq 3\/$ distinct vertices in $\/G\/$. An $x_{1}x_{2}\/$-hamiltonian path in $\/G^{2}\/$ which contains $k-2\/$ distinct edges $x_{i}y_{i}\in E(G)\/$, $\/i=3,\ldots,k\/$ is said to be $\/{\cal F}_{k}$. Hence we speak of an ${\cal F}_{k}$ $x_{1}x_{2}\/$-hamiltonian path. If $x_{i}\/$ is adjacent to $x_{j}\/$, we insist that $x_{i}y_{i}\/$ and $x_{j}y_{j}\/$ are distinct edges. A graph $\/G\/$ is said to have the ${\cal F}_{k}$ property if for any set $\/A=\{x_{1},\ldots,x_{k}\}\subseteq V(G)\/$, there is an ${\cal F}_{k}$ $x_{1}x_{2}\/$-hamiltonian path in $\/G^{2}\/$.

Let $G\/$ be a graph. By an $EPS\/$-graph, $JEPS\/$-graph respectively, of $G\/$, denoted $S=E\cup P\/$, $S=J\cup E\cup P\/$ respectively, we mean a spanning connected subgraph $S\/$ of $G\/$ which is the edge-disjoint union of an eulerian graph $E\/$ (which may be disconnected) and a linear forest $P\/$, respectively a linear forest $P\/$ together with an open trail $J\/$. For $S=E\cup P\/$, let $d_{E}(v)\/$ and $d_{P}(v)\/$ denote the degree of $v\/$ in $E\/$ and $P\/$, respectively. In the ensuing discussion we need, however, special types of $EPS$-graphs: thus a $[v;w]$-$EPS$-graph $S=E\cup P$ of $G$ with $v,w\in V(G)$, satisfies $d_{P}(v)=0$ and $d_{P}(w)\leq 1$. For $k\geq 2$, $[v;w_{1},\dots,w_{k}]$-$EPS$-graphs are defined analogously, whereas in $[w_{1}\dots,w_{k}]$-$EPS$-graphs only $d_{P}(w_{i})\leq 1$, $i=1,\dots,k$, needs to be satisfied.

Let $bc(G)\/$ denote the block-cutvertex graph of the graph $G\/$. If $bc(G)\/$ is a path, we call $G\/$ a block chain. A block chain $G\/$ is called trivial if $E(bc(G))=\emptyset\/$; otherwise it is called non-trivial. A block of $G\/$ is an endblock of $G\/$ if it contains at most one cutvertex of $G\/$.

In [5, Lemma 2], it was shown that if $G\/$ is a block chain whose endblocks $B_{1},B_{2}\/$ are $2\/$-connected and $v\in B_{1}\/$ and $w\in B_{2}\/$ are not cutvertices of $G\/$, then $G\/$ has an $EPS\/$-graph $S=E\cup P\/$ such that $d_{P}(v)=0=d_{P}(w)\/$. A more refined statement is now given below. In Lemma 1 we apply [5, Lemma 2, Theorem 3] and in Theorem 1 we apply Theorem D (stated explicitly below) several times to the blocks of $G$, respectively to $G$ itself, to obtain $EPS$-graphs of the required type.

Lemma 1

Suppose $G\/$ is a block chain with a cutvertex, $v\/$ and $w\/$ are vertices in different endblocks of $G\/$ and are not cutvertices. Then

(i) there exists an $EPS\/$-graph $E\cup P\subseteq G\/$ such that $d_{P}(v),\ d_{P}(w)\leq 1\/$. If the endblock which contains $v\/$ is $2\/$-connected, then we have $d_{P}(v)=0\/$ and $d_{P}(w)\leq 1\/$; and

(ii) there exists a $JEPS\/$-graph $J\cup E\cup P\subseteq G\/$ such that $d_{P}(v)=0=d_{P}(w)\/$. Moreover, $v,w\/$ are the only odd vertices of $J\/$. Also, we have $d_{P}(c)=2\/$ for at most one cutvertex $c\/$ of $G\/$ (and hence $d_{P}(c^{\prime})\leq 1\/$ for all other cutvertices $c^{\prime}\/$ of $G\/$).

Proof: If $G\/$ is a path, the result is trivially true.

So assume that $G\/$ is not a path. If $G\/$ has a suspended path (i.e., a maximal path whose internal vertices are 2-valent in $G$) starting at the endvertex $v\/$ of $G$, then let $P_{v}\/$ denote this path and let $v_{1}\/$ denote the other endvertex of $P_{v}\/$. Note that $v_{1}\/$ is a cutvertex of $G\/$. If there is no such suspended path, then define $P_{v}\/$ to be an empty path. Likewise, $P_{w}\/$ is defined similarly with $w\/$ (respectively $w_{1}\/$) taking the place of $v\/$ (respectively $v_{1}\/$).

(i) By [5, Lemma 2], $G^{\prime}=G-(P_{v}\cup P_{w})\/$ has an $EPS\/$-graph $S^{\prime}=E^{\prime}\cup P^{\prime}\/$ with $d_{P^{\prime}}(v_{1})=0\/$ and $d_{P^{\prime}}(w_{1})\leq 1\/$. But this means that $G\/$ has an $EPS\/$-graph $S=E\cup P\/$ with $d_{P}(v)\leq 1\/$ and $d_{P}(w)\leq 1\/$ if we set $E=E^{\prime}\/$ and $P=P^{\prime}\cup P_{v}\cup P_{w}\/$. Clearly, in the case that $P_{v}\/$ is an empty path, then $v=v_{1}\/$ and we have $d_{P}(v)=0\/$ and $d_{P}(w)\leq 1\/$.

(ii) Let $B\/$ be a block of $G\/$. Let $c_{1},c_{2}\in V(B)$. If $B\/$ is not an endblock, then let $c_{1},c_{2}\in B\/$ be the cutvertices of $G\/$ in $B\/$. If $B\/$ is an endblock of $G\/$, then let only one of $c_{1},c_{2}\/$, say $c_{2}\/$, to be a cutvertex of $G\/$, and let $c_{1}=v$, $c_{1}=w$ respectively, depending on the endblock $c_{1}$ belongs to. By [5, Theorem 3], $B\/$ has a $JEPS\/$-graph $S_{B}=J_{B}\cup E_{B}\cup P_{B}\/$ with $d_{P_{B}}(c_{1})=0\/$, $d_{P_{B}}(c_{2})\leq 1\/$, and $c_{1},c_{2}\/$ are the only odd vertices of $J_{B}\/$. If $B$ is not an endblock, then we may interchange $c_{1}$ and $c_{2}$. Thus we can ensure that for at most two blocks of $G$, $B^{\prime}$ and $B^{\prime\prime}$ say, satisfying $B^{\prime}\cap B^{\prime\prime}=c_{2}$, we have $d_{P_{B^{\prime}}}(c_{2})=d_{P_{B^{\prime\prime}}}(c_{2})=1$.

Note that if $B\/$ is not a $2\/$-connected block, then $E_{B}=\emptyset=P_{B}\/$ so that $S_{B}=J_{B}\/$. In this case, $d_{P_{B}}(c_{1})=0=d_{P_{B}}(c_{2})\/$.

By taking $S=\bigcup_{B}S_{B}\/$, where the union is taken over all blocks $B\/$ of $G\/$, we have a $JEPS\/$-graph that satisfies the conclusion of (ii).

This completes the proof.

Theorem 1

Suppose $G\/$ is a $2\/$-connected graph and $v,w\/$ are two distinct vertices in $G\/$. Then either

(i) there exists an $EPS\/$-graph $S=E\cup P\subseteq G\/$ with $d_{P}(v)=0=d_{P}(w)\/$;

or

(ii) there exists a $JEPS\/$-graph $S=J\cup E\cup P\subseteq G\/$ with $v,w\/$ being the only odd vertices of $J\/$, and $d_{P}(v)=0=d_{P}(w)\/$.

Proof: If $G\/$ is a cycle, then clearly the result is true. Hence assume that $G\/$ is not a cycle.

Let $K^{\prime}\/$ be a cycle in $G\/$ containing $v,w\/$. If $d_{G}(v)=2\/$, then we take a $[w;v]$-$EPS$-graph with $K^{\prime}\subseteq E\/$. If $d_{G}(w)=2\/$, then we take a $[v;w]$-$EPS$-graph with $K^{\prime}\subseteq E\/$. In either case, Theorem D (stated below) guarantees the existence of such $EPS$-graphs. Thus conclusion (i) of the theorem is satisfied.

Hence we assume that $d_{G}(v),d_{G}(w)\geq 3\/$. We proceed by contradiction, letting $G\/$ be a counterexample with minimum $|E(G)|\/$.

Let $G^{\prime}=G-K^{\prime}\/$ denote the graph obtained from $G\/$ by deleting all edges of $K^{\prime}\/$ (including all possibly resulting isolated vertices).

(a) Suppose $G^{\prime}\/$ is $2\/$-connected. $G^{\prime}\/$ either has an $EPS\/$-graph $S^{\prime}=E^{\prime}\cup P^{\prime}\/$ or a $JEPS\/$-graph $S^{\prime}=J^{\prime}\cup E^{\prime}\cup P^{\prime}\/$ satisfying the additional property (i) or (ii), respectively.

Suppose $S^{\prime}=E^{\prime}\cup P^{\prime}\/$. Then set $E=K^{\prime}\cup E^{\prime}$, $P=P^{\prime}\/$ to obtain an $EPS\/$-graph $S=E\cup P\/$ of $G\/$ satisfying property (i). If $G^{\prime}$ has a $JEPS$-graph $S^{\prime}=J^{\prime}\cup E^{\prime}\cup P^{\prime}$ satisfying property (ii), then set $E=E^{\prime}$, $P=P^{\prime}$ and $J=J^{\prime}\cup K^{\prime}$, to obtain a $JEPS$-graph $S=J\cup E\cup P$ as required. Whence $G^{\prime}$ is not $2$-connected.

(b) Suppose $G^{\prime}$ has an endblock $B^{\prime}$ with $(B^{\prime}-\gamma c^{\prime})\cap\{v,w\}=\emptyset$ where $\gamma c^{\prime}=c^{\prime}$ if $B^{\prime}$ contains a cutvertex $c^{\prime}$ of $G^{\prime}$, and $\gamma c^{\prime}=\emptyset$ otherwise (in this latter case, $B^{\prime}$ is a component of $G^{\prime}$ having at least two vertices with $K^{\prime}$ in common). It follows that $G^{\prime}\supseteq H^{\prime}$ where $H^{\prime}$ is a block chain with $B^{\prime}\subseteq H^{\prime}$ and $G^{*}:=G-H^{\prime}$ is $2$-connected. Suppose $H^{\prime}$ is chosen in such a way that $G^{*}$ is as large as possible.

It follows that if $H^{\prime}$ is not $2$-connected then $|V(G^{*})\cap V(H^{\prime}-V(B^{\prime}))|=1$. Denote the corresponding vertex with $c^{*}$ and observe that $c^{*}$ is a cutvertex if $c^{*}\in V(G^{\prime})$. Also, by the choice of $B^{\prime}$ and the maximality of $G^{*}$ we have

[TABLE]

and $c^{*}$ is not a cutvertex of $H^{\prime}$. Let $u^{\prime}\in V(B^{\prime})-\gamma c^{\prime}$ be chosen arbitrarily. We set $\delta c^{*}=c^{*}$ if $c^{*}$ is a pendant vertex of $H^{\prime}$, and $\delta c^{*}=\emptyset$ otherwise. By repeated application of Theorem D (see below) we obtain an $EPS$-graph $S^{\prime}=E^{\prime}\cup P^{\prime}$ of $H^{\prime}-\delta c^{*}\/$ with $d_{P^{\prime}}(\delta c^{*})=0$ (setting $d_{P^{\prime}}(\emptyset)=0$) and $d_{P^{\prime}}(u^{\prime})\leq 1$.

If however, $H^{\prime}$ is 2-connected, i.e. $H^{\prime}=B^{\prime}$, then we let $c^{*}=(G^{\prime}-B^{\prime})\cap B^{\prime}$, if $B^{\prime}$ contains a cutvertex of $G^{\prime}$, otherwise $c^{*}\in V(B^{\prime})\cap V(K^{\prime})$ arbitrarily. Futhermore we choose $u^{\prime}\in V(B^{\prime})-c^{*}$ arbitrarily. By Theorem D, $B^{\prime}=H^{\prime}$ has a $[c^{*};u^{\prime}]$-$EPS$-graph $S^{\prime}=E^{\prime}\cup P^{\prime}$.

Also, $G^{*}$ has an $EPS$-graph $S^{*}=E^{*}\cup P^{*}$ or a $JEPS$-graph $S^{*}=J^{*}\cup E^{*}\cup P^{*}$ with $d_{P^{*}}(v)=d_{P^{*}}(w)=0$; and $K^{\prime}\subset E^{*}$, $K^{\prime}\subset J^{*}\cup E^{*}$ respectively.

Observing that $P^{*}\cap P^{\prime}=\emptyset$ and that $S^{*}$ and $S^{\prime}$ are edge-disjoint, we conclude that $E=E^{*}\cup E^{\prime}$ and $P=P^{*}\cup P^{\prime}$ together with $J=J^{*}$ yield $S=E\cup P$, $S=J\cup E\cup P$ respectively, a spanning subgraph of $G$ as claimed by the theorem (observe that $d_{P}(c^{*})=d_{P^{*}}(c^{*})\/$ because $d_{P^{\prime}}(c^{*})=0\/$, and $d_{P^{*}}(c^{*})=0\/$ if $c^{*}\in\{v,w\}\/$).

(c) Because of the cases solved already, we now show that $G^{\prime}$ is connected and for every endblock $B^{\prime}$ of $G^{\prime}$, $V(B^{\prime})\cap\{v,w\}\not=\emptyset$. For, if $G^{\prime}$ is disconnected and because of case (b) already solved, $G^{\prime}$ could be written as

[TABLE]

where $G^{\prime}_{i}$ is a component of $G^{\prime}$; and

[TABLE]

Without loss of generality $v\in G^{\prime}_{1}$, $w\in G^{\prime}_{2}$. Consequently, $G_{i}:=G^{\prime}_{i}\cup K^{\prime}$, $i=1,2$, is $2$-connected with $d_{G_{1}}(w)=2$, $d_{G_{2}}(v)=2$. Arguing as at the very beginning of the proof of this theorem (where we considered the case $d_{G}(v)=2$ or $d_{G}(w)=2$) we conclude that the corresponding $EPS$-graphs $S_{i}=E_{i}\cup P_{i}$ with $K^{\prime}\subseteq E_{i}$, $i=1,2$, satisfy conclusion (i) of the theorem, and so does $S=E\cup P$ where $E=E_{1}\cup(E_{2}-K^{\prime})$ and $P=P_{1}\cup P_{2}$.

Because of case (a) already solved, we thus have that $G^{\prime}$ is a non-trivial block chain with $v,w$ belonging to different endblocks $B_{v},B_{w}$ respectively, of $G^{\prime}$ and they are not cutvertices of $G^{\prime}$. Let $c_{v}$ and $c_{w}$ be the respective cutvertices of $B_{v}$ and $B_{w}$ (possibly $c_{v}=c_{w}$). If $B_{v}$ is not a bridge of $G^{\prime}$ we use a $[v;c_{v}]$-$EPS$-graph $S_{v}$ of $B_{v}$ and a $[w;c_{w}]$-$EPS$-graph $S_{w}$ of $B_{w}$ if $B_{w}$ is also not a bridge, or $S_{w}=\emptyset$ if $B_{w}$ is a bridge. Proceeding similarly for every block $B$ of $G^{\prime}-(B_{v}\cup B_{w})$ we conclude that $G^{\prime}$ has an $EPS$-graph $S^{\prime}=E^{\prime}\cup P^{\prime}$ with $d_{P^{\prime}}(v)=d_{P_{v}}(v)=0$ and $d_{P^{\prime}}(w)=d_{P_{w}}(w)=0$, where $P_{v}\subseteq S_{v}$, $P_{w}\subseteq S_{w}$ (defining $d_{P_{w}}(w)=0\/$ if $P_{w}=\emptyset\/$). Thus in either case $S^{\prime}\cup K^{\prime}$ is an $EPS$-graph of $G$ satisfying conclusion (i). However, if both $B_{v}$ and $B_{w}$ are bridges, i.e., $d_{G^{\prime}}(v)=d_{G^{\prime}}(w)=1$, we introduce $z\notin V(G^{\prime})$ and form $G_{z}:=G^{\prime}\cup\{z,zv,zw\}$. $G_{z}$ contains a cycle $K_{z}$ through $z,v,w$ since $\kappa(G_{z})\geq 2$, so it contains a $[v,w]$-$EPS$-graph $S_{z}=E_{z}\cup P_{z}$ with $K_{z}\subseteq E_{z}$. Trivially, $d_{P_{z}}(v)=d_{P_{z}}(w)=0$, and for the component $E_{0}\subseteq E_{z}$ with $z\in E_{0}$ we have $J:=(E_{0}-z)\cup K$ being an open trail joining $v$ and $w$. Setting $E=E_{z}-E_{0}$ and $P=P_{z}$ we conclude that $S=J\cup E\cup P$ is a $JEPS$-graph satisfying conclusion (ii) of the theorem. Theorem 1 now follows.

The following results from [8], [5], and [7] will be used quite frequently in the proof of Theorem 2.

Let $G\/$ be a graph and let $W\/$ be a set of vertices in $G\/$. A cycle $K\/$ in $G\/$ is said to be $W$-maximal if $|V(K^{\prime})\cap W|\leq|V(K)\cap W|\/$ for any cycle $K^{\prime}\/$ of $G\/$. Moreover, we say that the $W$-maximal $K\/$* is $W$-sound* if $|V(K)\cap W|\geq 4\/$.

The following Theorems A and B are special cases of the theorems quoted.

Theorem A

([8, Theorem 4]) Let $G\/$ be a $2\/$-connected graph and let $W\/$ be a set of five distinct vertices in $G\/$. Suppose $K\/$ is a $W\/$-sound cycle in $G\/$. Then there is an $EPS\/$-graph $S=E\cup P\/$ of $G\/$ such that $K\subseteq E\/$ and $d_{P}(w)\leq 1\/$ for every $w\in W\/$.

An $EPS\/$-graph which satisfies the conclusion of Theorem A is also called a $W$-EPS-graph.

Theorem B

([8, Theorem 3]) Let $G\/$ be a $2\/$-connected graph and let $v,w_{1},w_{2},w_{3}\/$ be four distinct vertices of $G\/$. Suppose $K\/$ is a cycle in $G\/$ such that $\{v,w_{1},w_{2},w_{3}\}\subseteq K\/$. Then $G\/$ has a $[v;w_{1},w_{2},w_{3}]\/$-$EPS\/$-graph $S=E\cup P\/$ such that $K\subseteq E\/$.

Suppose $G\/$ is a $2\/$-connected graph and $v,w_{1},w_{2}\/$ are distinct vertices in $G\/$. A cycle $K\/$ in $G\/$ is a $[v;w_{1},w_{2}]\/$*-maximal cycle

• in $G\/$ if $\{v,w_{1}\}\subseteq V(K)\/$, and $w_{2}\in V(K)\/$ unless $G\/$ has no cycle containing all of $\{v,w_{1},w_{2}\}\/$.

Theorem C

([8, Theorem 2]) Let $G\/$ be a $2\/$-connected graph and let $v,w_{1},w_{2}\/$ be three distinct vertices of $G\/$. Suppose $K\/$ is a $[v;w_{1},w_{2}]\/$-maximal cycle in $G\/$. Then $G\/$ has a $[v;w_{1},w_{2}]\/$-$EPS\/$-graph $S=E\cup P\/$ such that $K\subseteq E\/$.

Theorem D

([5, Theorem 2]) Let $G\/$ be a $2\/$-connected graph and let $v,w\/$ be two distinct vertices of $G\/$. Let $K\/$ be a cycle through $v,w\/$. Then $G\/$ has a $[v;w]\/$-$EPS\/$-graph $S=E\cup P\/$ with $K\subseteq E\/$.

Theorem E

([7, Theorem 3]). Suppose $\/v\/$ and $\/w\/$ are two arbitrarily chosen vertices of a $\/2\/$-connected graph $\/G\/$. Then $\/G^{2}\/$ contains a hamiltonian cycle $C\/$ such that the edges of $\/C\/$ incident to $\/v\/$ are in $\/G\/$ and at least one of the edges of $\/C\/$ incident to $\/w\/$ is in $\/G\/$. Further, if $\/v\/$ and $\/w\/$ are adjacent in $\/G\/$, then these are three different edges.

A hamiltonian cycle in $G^{2}\/$ satisfying the conclusion of Theorem E is also called a $[v;w]\/$-hamiltonian cycle. More generally, a hamiltonian cycle $C\/$ in $G^{2}\/$ which contains two edges of $G\/$ incident to $v\/$, and at least one edge of $G\/$ incident to each $w_{i}\/$, $i=1,\ldots,k\/$, is called a $[v;w_{1},\ldots,w_{k}]\/$-hamiltonian cycle, provided the edges in question are all different.

Theorem F

([7, Theorem 4]). Let $G\/$ be a $\/2\/$-connected graph. Then the following hold.

(i) $G\/$ has the ${\cal F}_{3}\/$ property.

(ii) For a given $q\in\{x,y\}\/$, $G^{2}\/$ has an $xy\/$-hamiltonian path containing an edge of $G\/$ incident to $q\/$.

By applying Theorems E and F to each block of a block chain $B\/$, we have the following.

Corollary 1

Suppose $B\/$ is a non-trivial block chain with $|V(B)|\geq 3\/$ and $v\/$ and $w\/$ are vertices in different endblocks of $G\/$. Assume further that $v,w\/$ are not cutvertices of $B\/$. Then

(i) $B^{2}\/$ has a hamiltonian cycle which contains an edge of $B\/$ incident to $v\/$ and an edge of $B\/$ incident to $w\/$. In the case that the endblock which contains $v\/$ is $2\/$-connected, then $B^{2}$ has a hamiltonian cycle which contains two edges of $B$ incident to $v\/$ and an edge of $B\/$ incident to $w\/$. Also,

(ii) $B^{2}\/$ has a $vw\/$-hamiltonian path containing an edge of $B\/$ incident to $v\/$ and an edge of $B\/$ incident to $w\/$.

3 $DT\/$-graphs

Recall that a graph is called a $DT\/$-graph if every edge is incident to a $2\/$-valent vertex. If $G\/$ is a graph, we denote by $V_{2}(G)\/$ the set of all vertices of degree $2\/$ in $G\/$.

The following result which is interesting in itself, is obtained by applying Theorem 1 and the construction in [5] of a hamiltonian cycle/path in the corresponding spanning subgraph.

Corollary 2

Let $G\/$ be a $DT\/$-block and $x_{1},x_{2}\in V(G)$ satisfying $N(x_{1}),N(x_{2})\subseteq V_{2}(G)\/$ and $x_{1}x_{2}\not\in E(G)$. Then either (i) there exists a hamiltonian cycle in $G^{2}-x_{2}\/$ whose edges incident to $x_{1}\/$ are in $G\/$, or else (ii) there exists an $x_{1}x_{2}\/$-hamiltonian path in $G^{2}\/$ whose first and final edges are in $G\/$.

Theorem 2

Every $\/2\/$-connected $DT\/$-graph has the ${\cal F}_{4}$ property.

The proof of Theorem 2 is rather involved. We first give an outline of the general strategy used in the proof.

Let $G\/$ be a $2\/$-connected $DT\/$-graph and let $A=\{x_{1},x_{2},x_{3},x_{4}\}\/$ be a set of four distinct vertices in $G\/$. Let $G^{+}\/$ denote the $2\/$-connected graph obtained from $G\/$ by adding a new vertex $y\/$ which joins $x_{1}\/$ and $x_{2}\/$. Then $G^{+}\/$ is a $DT\/$-graph unless $N_{G}(x_{i})\not\subseteq V_{2}(G)\/$ for some $i\in\{1,2\}\/$. We shall show that $(G^{+})^{2}\/$ contains a hamiltonian cycle $C\/$ containing edges of $G^{+}$ of the form $yx_{1},yx_{2},x_{3}z_{3},x_{4}z_{4}\/$ where $x_{3}z_{3},x_{4}z_{4}\/$ are edges of $G\/$. Then clearly $C\/$ gives rise to an ${\cal F}_{4}$ $x_{1}x_{2}\/$-hamiltonian path in $G^{2}\/$ when we delete the vertex $y\/$ from $(G^{+})^{2}\/$.

In order to show the existence of such hamiltonian cycle $C\/$ in $(G^{+})^{2}\/$, we shall apply induction or show that $G^{+}\/$ admits an $EPS\/$-graph $S=E\cup P\/$ with some additional properties. In particular, in almost all cases, $E\/$ will contain a prescribed cycle $K^{+}\/$ passing through $y\/$. $K^{+}\/$ will also contain as many elements of $\{x_{3},x_{4}\}\/$ as possible. Note that $G^{+}$ is $2$-connected and hence contains a cycle through $y$ and $x_{i}$, $i\in\{3,4\}$, which automatically contains $x_{1},x_{2}$.

Note that in [5] it was shown that if a $2\/$-connected $DT\/$-graph $H\/$ admits an $EPS\/$-graph, then $H^{2}\/$ has a hamiltonian cycle. We refer the reader to [5] for the method of constructing such hamiltonian cycle and to see how edges of $H\/$ can be included in such hamiltonian cycle. Also, we may automatically assume that in an $EPS\/$-graph $S=E\cup P\/$ the edges of $P\/$ are the bridges of $S\/$ (otherwise, we could delete step-by-step $P\/$-edges (i.e., edges of $P\/$) until such situation is achieved).

However, $G^{+}\/$ may not be a $DT\/$-graph and/or some elements in $A\/$ may be $2\/$-valent and (at least) one of its neighbors may not be $2\/$-valent. In such cases, the existence of the various types of $EPS\/$-graphs $S\/$ in $G^{+}\/$ may not be sufficient to guarantee a hamiltonian cycle to begin with in $S^{2}\/$. Even if we can derive the existence of a hamiltonian cycle from these $EPS$-graphs, they may not suffice to guarantee a hamiltonian cycle with the additional properties. Thus we need to consider neighbors of elements of $A\/$ to assure that they are incident to less than two $P\/$-edges. This applies, in particular, to $z_{i}\in N_{G}(x_{i})\/$ with $z_{i}x_{i}\in E(K^{+})$, $i\in\{1,2,3,4\}\/$.

The following observations will be used quite frequently (sometimes implicitly) in the proof of Theorem 2.

Observation (*): Suppose $S=E\cup P\/$ is an $EPS\/$-graph of $G^{+}\/$ such that $d_{P}(x_{i})\leq 1\/$ for $i=1,2\/$. Let $x\/$ be a $2\/$-valent vertex of $G\/$ belonging to $E\/$.

(i) Suppose $N(x)=\{u_{1},u_{2}\}\/$. Then $S^{2}\/$ has a hamiltonian cycle which contains the edges $yx_{1},yx_{2}$ and $u_{i}x\/$ for some $i\in\{1,2\}$ unless $x_{j}\in N(u_{j})\cup\{u_{j}\}\/$ and $d_{P}(x_{j})=1,d_{S}(u_{j})>2$ for $j=1,2$; or for some $j\in\{1,2\}$, $d_{P}(x_{j})=1,d_{P}(z_{j})=2$ and $z_{j}\in N(x_{j})\cap V(K^{+})$; or $d_{P}(u_{1})=d_{P}(u_{2})=2$ - in all three cases $N_{G}(x_{j})\not\subseteq V_{2}(G)$.

(ii) We further note that any pendant edge in $S\/$ will always be contained in any hamiltonian cycle of $S^{2}\/$.

*(iii) Consider $W\subseteq V(G^{+})$ with $|W|=5$ and $K^{+}\subset G^{+}$. Suppose $|W\cap V(K^{+})|\geq 4$. If $K^{+}$ is $W$-sound, then Theorem A applies. If, however, $K^{+}$ is not $W$-sound, then there is a $W$-sound cycle $K^{*}\/$ with $W\subseteq K^{*}\/$ and we operate with $K^{*}\/$ in place of $K^{+}\/$. This follows from the definition of $W\/$-soundness (see the discussion immediately preceding Theorem A). *

The observations (i) and (ii) follow directly from the degree of freedom inherent in the construction of a hamiltonian cycle in $S^{2}\/$ as given in [5].

The proof of Theorem 2 is divided into several cases depending on whether $N(x_{i})\subseteq V_{2}(G)\/$ or not, $i=1,2,3,4$. Note that if $N(x_{i})\not\subseteq V_{2}(G)\/$, then $d_{G}(x_{i})=2\/$. If $d_{G}(x_{i})=2\/$, we let $N(x_{i})=\{u_{i},v_{i}\}\/$ throughout the proof. Also, we define $x_{i}^{*}=x_{i}\/$ if $d_{G}(x_{i})>2\/$; and $x_{i}^{*}=z_{i}$ otherwise.

Lemma 2

Let $G^{+}\/$ be defined as before with $N(x_{3})\not\subseteq V_{2}(G)\/$ and $N(x_{4})\not\subseteq V_{2}(G)\/$. Suppose $N(x_{i})\subseteq V_{2}(G)\/$ for some $i\in\{1,2\}\/$. Assume further that every proper $2\/$-connected subgraph of $G\/$ has the ${\cal F}_{4}$ property. Then $(G^{+})^{2}\/$ has a hamiltonian cycle containing the edges $x_{1}y,x_{2}y,x_{3}z_{3},x_{4}z_{4}\/$ where $x_{3}z_{3},x_{4}z_{4}\/$ are different edges of $G\/$.

Proof:   By the hypotheses, $d_{G}(x_{3})=d_{G}(x_{4})=2$. Assume without loss of generality that $N(x_{1})\subseteq V_{2}(G)\/$.

(1) Suppose $\{u_{i},v_{i}\}\neq\{x_{1},x_{2}\}\/$ for $i=3,4\/$.

Let $K^{+}\/$ be a cycle containing the vertices $y,x_{1},x_{2},x_{4},u_{4},v_{4}\/$.

(1.1) Assume that $K^{+}\/$ also contains the vertex $x_{3}\/$.

We may assume that

[TABLE]

(a) Assume that $u_{4}\neq x_{1}\/$.

Since $\{x_{1},x_{2},x_{3},x_{4},u_{3},u_{4},z_{2}\}\subseteq V(K^{+})\/$, Theorem B ensures the existence of a $[u_{4};x_{1},z_{2},x_{2}]\/$-$EPS\/$-graph $S_{4}=E_{4}\cup P_{4}\/$ of $G^{+}$ with $K^{+}\subseteq E_{4}\/$ in the case $x_{3}x_{4}\in E(G)\/$. Likewise, we obtain a $[u_{4};x_{1},u_{3},x_{2}^{*}]$-$EPS\/$-graph $S_{3}=E_{3}\cup P_{3}\/$ of $G^{+}$ with $K^{+}\subseteq E_{3}\/$ if $x_{3}x_{4}\not\in E(G)\/$ where $x_{2}^{*}=x_{2}\/$ if $d_{G}(x_{2})>2\/$, and $x_{2}^{*}=z_{2}=V(K^{+})\cap N_{G}(x_{2})\/$ otherwise. It is straightforward to see that in both cases, the $EPS\/$-graph yields a hamiltonian cycle in $(G^{+})^{2}\/$ as required by the lemma (see Observation (*)(i)).

(b) Assume that $u_{4}=x_{1}\/$ and $v_{3}=x_{2}\/$.

(b1) Suppose $x_{3}\/$ and $x_{4}\/$ are adjacent or $N(x_{3})\cap N(x_{4})\neq\emptyset\/$.

(i) $x_{3}\/$ and $x_{4}\/$ are adjacent. Let $G^{-}=G-\{x_{3},x_{4}\}\/$. If $G^{-}\/$ is not $2\/$-connected, then it is a non-trivial block chain with $x_{1},x_{2}\/$ belonging to different endblocks, and $x_{1},x_{2}\/$ are not cutvertices of $G^{-}\/$. Hence $(G^{-})^{2}\/$ has a hamiltonian path $P(x_{1},x_{2})\/$ starting with an edge $x_{1}w_{1}\/$ of $G\/$ and ending with an edge $x_{2}w_{2}\/$ of $G\/$ (see Corollary 1(ii)). Then

[TABLE]

defines a required ${\cal F}_{4}\/$ $x_{1}x_{2}\/$-hamiltonian path in $G^{2}\/$.

If $G^{-}\/$ is $2\/$-connected, then $(G^{-})^{2}\/$ has a hamiltonian cycle $C^{-}\/$ containing $x_{1}w_{1},x_{1}t_{1},\linebreak x_{2}w_{2}\/$ which are edges of $G\/$. Then

[TABLE]

is a required ${\cal F}_{4}\/$ $x_{1}x_{2}\/$-hamiltonian path in $G^{2}\/$.

(ii) Suppose $N(x_{3})\cap N(x_{4})=\{u\}\/$.

If $d_{G}(u)=2\/$, then let $G^{-}=G-\{x_{3},x_{4},u\}\/$ and proceed similarly as before to obtain a required ${\cal F}_{4}\/$ $x_{1}x_{2}\/$-hamiltonian path in $G^{2}\/$. Hence we assume that $d_{G}(u)>2\/$. Suppose further that $G-x_{i}\/$ is $2\/$-connected for some $i\in\{3,4\}\/$. Then $G-x_{i}\/$ has the ${\cal F}_{4}\/$ property with $u\/$ taking the place of $x_{i}\/$; and any such ${\cal F}_{4}\/$ $x_{1}x_{2}\/$-hamiltonian path in $(G-x_{i})^{2}\/$ can be extended to a required ${\cal F}_{4}\/$ $x_{1}x_{2}\/$-hamiltonian path in $G^{2}\/$. Thus we have to consider the case $\kappa(G-x_{i})<2\/$ for $i\in\{3,4\}\/$.

Consider $G-x_{4}\/$. Since $d_{G}(x_{4})=2\/$, $G^{\prime}=G-x_{4}\/$ is a non-trivial block chain with $x_{1},u\/$ belonging to different endblocks of $G^{\prime}\/$ and are not cutvertices of $G^{\prime}\/$. The endblock $B_{u}\/$ of $G^{\prime}\/$ with $u\in V(B_{u})\/$ also contains $x_{3},x_{2}\/$ because $d_{G^{\prime}}(u)\geq 2\/$ and $d_{G}(x_{3})=d_{G^{\prime}}(x_{3})=2\/$. Hence $B_{u}\/$ is $2\/$-connected. Let $c\/$ be the cutvertex of $G^{\prime}\/$ belonging to $B_{u}\/$.

Suppose first $c\neq x_{2}\/$. Because of the hypothesis of the lemma, $B_{u}\/$ has the ${\cal F}_{4}\/$ property. Correspondingly, there is a hamiltonian path $P(c,x_{2})\/$ in $(B_{u})^{2}\/$ containing $x_{3}w_{3},uu^{\prime}\/$ with $w_{3}\in\{u,x_{2}\}\/$, which are different edges of $B_{u}\/$. Likewise, there is a hamiltonian path $P(x_{1},c)\/$ in $(G^{\prime}-B_{u})^{2}\/$ by Theorem F, Corollary 1(ii), respectively. Then

[TABLE]

is a required ${\cal F}_{4}\/$ $x_{1}x_{2}\/$-hamiltonian path in $G^{2}\/$.

Finally suppose $c=x_{2}\/$. By Theorem F(ii) or Corollary 1(ii), $(G^{\prime}-B_{u})^{2}\/$ has a $x_{1}x_{2}\/$-hamiltonian path $P_{1,2}\/$ ending with an edge $w_{2}x_{2}\/$ of $G\/$. By Theorem E, $(B_{u})^{2}\/$ has a hamiltonian cycle $C_{u}\/$ with $\{ux_{3},x_{2}x_{3},z_{2}x_{2}\}\subset E(B_{u})\/$.

[TABLE]

defines a hamiltonian path as required.

(b2) Suppose $x_{3}\/$ and $x_{4}\/$ are not adjacent and $N(x_{3})\cap N(x_{4})=\emptyset\/$.

Let $W=\{y,x_{1},x_{2},u_{3},v_{4}\}\/$. Then $K^{+}\/$ is $W\/$-sound. By Theorem A, $G^{+}\/$ has an $EPS\/$-graph $S=E\cup P\/$ with $K^{+}\subseteq E\/$ and $d_{P}(w)\leq 1\/$ for every $w\in W\/$; and $d_{P}(x_{3})=d_{P}(x_{4})=0$. Because of the hypothesis of this case a required hamiltonian cycle can be constructed in $(G^{+})^{2}\/$ (see Observation (*)(i)). In particular, the hamiltonian cycle contains $x_{4}v_{4}$ and $u_{3}x_{3}$.

(c) Assume that $u_{4}=x_{1}\/$ and $v_{3}\neq x_{2}$.

If $x_{3}x_{4}\not\in E(K^{+})\/$, then Theorem B ensures the existence of an $[x_{2};x_{1},v_{4},v_{3}]$-$EPS\/$-graph $S_{3}=E_{3}\cup P_{3}\/$ of $G^{+}$ with $K^{+}\subseteq E_{3}\/$. By construction, $S^{2}\/$ contains a hamiltonian cycle $C\/$ with $x_{4}v_{4},x_{3}v_{3}\in E(C)\/$. (see Observation (*)(i)). Hence we assume that $x_{3}x_{4}\in E(K^{+})\/$.

If $v_{3}x_{2}\not\in E(K^{+})\/$, or $v_{3}x_{2}\in E(K^{+})\/$ and $d_{G}(x_{2})>2\/$, then we invoke Theorem C to obtain a $[v_{3};x_{1},x_{2}^{*}]$-$EPS\/$-graph $S_{3}=E_{3}\cup P_{3}\/$ of $G^{+}$ with $K^{+}\subseteq E_{3}\/$. If, however, $v_{3}x_{2}\in E(K^{+})\/$ and $d_{G}(x_{2})=2\/$, then Theorem C ensures the existence of an $[x_{2};x_{1},v_{3}]$-$EPS\/$-graph $S_{3}=E_{3}\cup P_{3}\/$ of $G^{+}$ with $K^{+}\subseteq E_{3}\/$. Note that $K^{+}\/$ contains all these special vertices. In all these cases, $(S_{3})^{2}\/$ contains a hamiltonian cycle $C\/$ with $x_{3}x_{4},x_{3}v_{3}\in E(C)\/$ (see Observation (*)(i)).

(1.2) In view of case (1.1), we may assume that $G^{+}\/$ has no cycle containing $y,x_{4},x_{3}\/$, and that

[TABLE]

and $G^{+}-x_{3}$ is 2-connected if $G^{+}-x_{i}$ is 2-connected for some $i\in\{3,4\}$.

Without loss of generality, assume that $u_{3}\not\in\{x_{1},x_{2}\}\/$.

(a) Consider first the case that $G^{*}=G^{+}-x_{3}\/$ is $2\/$-connected.

Define $W^{*}=\{y,x_{1},x_{2}^{*},u_{4},u_{3}\}\/$ if $x_{1}\neq u_{4}\/$ and $W^{*}=\{y,x_{1},x_{2}^{*},v_{4},u_{3}\}\/$ otherwise. Abbreviate $W^{*}=\{y,x_{1},x_{2}^{*},t_{4},u_{3}\}\/$ with $t_{4}\in\{u_{4},v_{4}\}\/$.

(a1) We first deal with the case $|W^{*}|=5\/$.

In view of Observation (*)(iii), set $K^{*}=K^{+}\/$ if $K^{+}\/$ is $W^{*}\/$-sound in $G^{*}$, or else there exists $K^{*}\supset W^{*}\/$ in $G^{*}$ (note $|K^{+}\cap W^{*}|\geq 4\/$).

(a1.1) Assume that $x_{4}\in K^{*}$. In this case we may assume that $K^{+}=K^{*}$. By Theorem A, there exists a $W^{*}$-$EPS$-graph $S^{*}=E^{*}\cup P^{*}$ of $G^{*}$ with $K^{*}\subseteq E^{*}$. Noting that $d_{P^{*}}(u_{3})\leq 1$, we set $E=E^{*}$ and $P=P^{*}\cup\{u_{3}x_{3}\}$. Then $S=E\cup P\/$ is an $EPS\/$-graph of $G^{+}\/$ whose structure implies that $(G^{+})^{2}\/$ has a hamiltonian cycle containing the edges $u_{3}x_{3}\/$ and $t_{4}x\/$ (because $x_{3}\/$ is a pendant vertex in $S\/$ - see Observation (*)(i)-(ii)).

(a1.2) Assume that $x_{4}\not\in K^{*}\/$. Then $u_{3}\in K^{*}$ (hence $K^{+}\neq K^{*}$). Since $x_{i}\notin K^{*}$, for $i=3,4$, $d_{G}(t_{4})>2$, $d_{G}(u_{3})>2$. We define $x^{**}_{2}$ as $x^{*}_{2}$ with respect to $K^{*}$.

First suppose $x^{**}_{2}=x^{*}_{2}$. By Theorem B, there exists a $[u_{4};x_{1},u_{3},x_{2}^{*}]$-$EPS\/$-graph $S^{*}=E^{*}\cup P^{*}\/$ of $G^{*}\/$ with $K^{*}\subseteq E^{*}\/$ if $x_{1}\neq u_{4}\/$. By the same token, there is a $[u_{4};v_{4},u_{3},x_{2}^{*}]$-$EPS\/$-graph $S^{*}=E^{*}\cup P^{*}\/$ of $G^{*}\/$ with $K^{*}\subseteq E^{*}\/$ if $x_{1}=u_{4}\/$. In both cases, we set $E=E^{*}\/$, $P=P^{*}\cup\{x_{3}u_{3}\}\/$. Then $S=E\cup P\/$ is an $EPS\/$-graph of $G^{+}\/$ which yields a hamiltonian cycle in $(G^{+})^{2}\/$ containing $u_{3}x_{3}\/$ and $x_{4}z\/$ for some $z\in N(x_{4})\/$. If $x_{4}\/$ is a pendant vertex in $S^{*}\/$, then it is adjacent to $v_{4}\/$ (see Observation (*)(i)-(ii)).

If $x^{**}_{2}\neq x^{*}_{2}$, then we proceed analogously as before using $x^{**}_{2}$ instead of $x^{*}_{2}$. Note that $u_{3}=x^{**}_{2}$ is not an obstacle (we use Theorem C) because of $d_{S}(x_{2})=2$ since $d_{G}(x_{2})=2$ and $x_{2}^{*}$ is also in $K^{*}$ (thus $x_{2}x_{2}^{*}\notin E(S)$) in this case.

(a2) Assume that $|W^{*}|=4.\/$

(a2.1) $W^{*}=\{y,x_{1},x_{2}^{*},t_{4}\}$ where $t_{4}\in\{u_{4},v_{4}\}$. If $u_{3}=t_{4}$, then we operate with a $[t_{4};x_{1},x_{2}^{*}]$-EPS graph $S^{*}=E^{*}\cup P^{*}$ of $G^{*}\/$ with $K^{+}\subseteq E^{*}$, which exists by Theorem C. If $u_{3}=x_{2}^{*}$, then we operate with a $[x_{2}^{*};x_{1},t_{4}]$-EPS graph $S^{*}=E^{*}\cup P^{*}$ of $G^{*}\/$ with $K^{+}\subseteq E^{*}$, which exists by Theorem C (note that $x_{2}^{*}\neq x_{2}$ in this case using $u_{3}\neq x_{2}$).

In either case, set $E=E^{*}\/$ and $P=P^{*}\cup\{x_{3}u_{3}\}\/$. Then $S=E\cup P\/$ is an $EPS\/$-graph of $G^{+}\/$ which yields a hamiltonian cycle $C\/$ in $(G^{+})^{2}\/$ containing either $x_{3}t_{4}x_{4}\/$ or $x_{2}^{*}x_{3}\/$, $x_{4}t_{4}\/$ (see Observation (*)(i)).

(a2.2) $W^{*}=\{y,x_{1},x_{2}^{*},u_{3}\}\/$. Then either (i) $u_{4}\neq x_{1}\/$, or (ii) $u_{4}=x_{1}\/$ and $v_{4}\neq x_{2}^{*}\/$ or (iii) $u_{4}=x_{1}\/$ and $v_{4}=x_{2}^{*}\/$.

In cases (i) and (ii) we are back to case (a2.1) with $u_{3}=t_{4}\neq x_{2}^{*}$.

In case (iii) we have $x_{2}^{*}\neq x_{2}\/$ because $N(x_{4})\neq\{x_{1},x_{2}\}\/$. We consider $G^{\prime}=G^{+}-\{x_{4},\delta x_{2}^{*}\}\/$; again, $\delta x_{2}^{*}=x_{2}^{*}\/$ if $x_{2}^{*}\/$ is a pendant vertex in $G^{+}-x_{4}\/$ and $\delta x_{2}^{*}=\emptyset\/$ otherwise. Set $x_{2}^{\prime}=x_{2}^{*}\/$ if $x_{2}^{*}\in V(G^{\prime})\/$ and $x_{2}^{\prime}=x_{2}\/$ otherwise. Suppose $\kappa(G^{\prime})=1\/$. In any case, $G^{\prime}\/$ has different endblocks $B_{1}^{\prime}\/$ and $B_{2}^{\prime}\/$; they are $2\/$-connected with $x_{1}\in B_{1}^{\prime}\/$ and $x_{2}^{\prime}\in B_{2}^{\prime}\/$ not being cutvertices of $G^{\prime}\/$. Since $G^{\prime}\/$ is homeomorphic to $G\/$ if $x_{2}^{\prime}=x_{2}\/$ (a contradiction to $\kappa(G^{\prime})=1\/$), it follows that $x_{2}^{*}\in B_{2}^{\prime}\/$ and that $x_{2}\/$ is a cutvertex of $G^{\prime}\/$ since $\{x_{2}\}=B_{1}^{\prime}\cap B_{2}^{\prime}\/$. However, $3=d_{G^{+}}(x_{2})=d_{B_{1}^{\prime}}(x_{2})+d_{B_{2}^{\prime}}(x_{2})\geq 2+2\/$, an obvious contradiction. Thus $G^{\prime}\/$ is $2\/$-connected in any case. Starting with a cycle $K^{\prime}\subseteq G^{\prime}\/$ with $y,x_{1},x_{2},x_{3}\in V(K^{\prime})\/$ we apply Theorem C to obtain an $[x_{1};u_{3},x_{2}^{**}]$-$EPS\/$-graph $S^{\prime}=E^{\prime}\cup P^{\prime}\/$ of $G^{\prime}\/$ with $K^{\prime}\subseteq E^{\prime}\/$, where $x_{2}^{**}=N_{G}(x_{2})-x_{2}^{*}$. Setting $E=E^{\prime}\/$, $P=P^{\prime}\cup\{x_{1}x_{4},\delta(x_{4}x_{2}^{*})\}\/$, where $\delta(x_{4}x_{2}^{*})=x_{4}x_{2}^{*}$ if $x_{2}^{*}\notin V(G^{\prime})$ and $\delta(x_{4}x_{2}^{*})=\emptyset$ otherwise, we obtain $S=E\cup P\/$ of $G^{+}$ with $K^{\prime}\subseteq E\/$ and $d_{P}(x_{1})=1\/$ and $d_{P}(u_{3})\leq 1\/$. It is clear that $S^{2}\/$ yields a hamiltonian cycle of $(G^{+})^{2}\/$ as required (see Observation (*)(i)).

(a3) Assume that $|W^{*}|=3.\/$

Then $W^{*}=\{y,x_{1},x_{2}^{*}\}\/$.

Hence $u_{3}\notin\{x_{1},y\}$, therefore $u_{3}=x_{2}^{*}$. Analogously $t_{4}\notin\{x_{1},y\}$, therefore $t_{4}=v_{4}=x_{2}^{*}$. That is, $u_{3}=x_{2}^{*}=t_{4}=v_{4}$ and $x_{1}=u_{4}$. $G^{\prime}=G^{+}-x_{4}\/$ is $2\/$-connected since there is a cycle $K^{\prime}\/$ in $G^{\prime}\/$ containing $y\/$ and $x_{3}\/$ and hence also $x_{2}^{*},x_{1},v_{3}\/$. If $v_{3}\neq x_{1}\/$, we operate with an $[x_{2}^{*};x_{1},v_{3}]$-$EPS\/$-graph $S^{\prime}=E^{\prime}\cup P^{\prime}\/$ of $G^{\prime}\/$ with $K^{\prime}\subseteq E^{*}\/$ (by Theorem C). Setting $E=E^{*}\/$ and $P=P^{*}\cup\{x_{4}x_{2}^{*}\}\/$, we obtain an $EPS\/$-graph $S=E\cup P\/$ of $G^{+}$ which will yield a hamiltonian cycle in $(G^{+})^{2}\/$ containing $x_{3}v_{3},x_{4}v_{4}\/$ (see Observation (*)(i)). If $v_{3}=x_{1}\/$, then $G-x_{4}\/$ is $2\/$-connected (since $N(x_{3})=N(x_{4})\/$). Hence $G-x_{4}\/$ has the ${\cal F}_{4}\/$ property with $v_{4}\/$ taking the place of $x_{4}\/$; and any such ${\cal F}_{4}\/$ $x_{1}x_{2}\/$-hamiltonian path in $(G-x_{4})^{2}\/$ can be extended to a required ${\cal F}_{4}\/$ $x_{1}x_{2}\/$-hamiltonian path in $G^{2}\/$. This finishes the proof of case (a).

(b)  Now consider the case where $G^{*}=G^{+}-x_{3}\/$ has a cutvertex and hence $G^{+}-x_{4}$ has also a cuvertex, because of the assumptions of case (1.2). Thus $G^{*}$ is a non-trivial block chain since $d_{G}(x_{3})=2$. Note that $K^{+}\/$ is contained in some endblock $B_{y}\/$ of $G^{*}\/$.

Let $W=\{y,x_{1},x_{2}^{*},x_{3},t_{4}\}\/$ where we define $t_{4}$ as follows:

• •

$t_{4}=u_{4}$ if $u_{4}\neq x_{1}$;

• •

$t_{4}=v_{4}$ if $u_{4}=x_{1}$ and either $x_{2}^{*}=x_{2}$ or $v_{4}\neq x_{2}^{*}\neq x_{2}$;

• •

$t_{4}=x_{2}$ if $u_{4}=x_{1}$ and $v_{4}=x_{2}^{*}\neq x_{2}$.

Note that by this definition of $t_{4}$, $|W|=5$.

Assume first that the cycle $K^{+}\/$ (which passes through $y,x_{1},x_{2},x_{2}^{*},u_{4},x_{4},v_{4}\/$) is $W$-sound in $G^{+}$. Let $\widehat{G}\/$ denote the subgraph of $G^{+}\/$ which is a non-trivial block chain containing $u_{3},x_{3},v_{3}\/$ such that $G^{+}-\widehat{G}=B_{y}\/$. Suppose $w_{3}\/$ is the vertex in one of the endblocks of $\widehat{G}$ and $w^{\prime}_{3}$ the vertex in the other endblock of $\widehat{G}$ such that $\widehat{G}\cap B_{y}=\{w_{3},w^{\prime}_{3}\}$. Possibly $\{w_{3},w^{\prime}_{3}\}\cap\{u_{3},v_{3}\}\neq\emptyset\/$, but $\{w_{3},w^{\prime}_{3}\}\neq\{u_{3},v_{3}\}\/$.

We replace $\widehat{G}\/$ in $G^{+}\/$ by a path $P_{4}=a_{1}a_{2}x_{3}a_{3}a_{4}\/$ (where $a_{1},a_{3}\/$ are identified with $w_{3},w^{\prime}_{3}\/$ respectively, and $\{a_{2},a_{3}\}=\{u_{3},v_{3}\}$) to obtain the graph $G^{\prime\prime}\/$. Note that $K^{+}\subseteq G^{\prime\prime}\/$. Set $W=\{y,x_{1},x_{2}^{*},x_{3},t_{4}\}\/$ as above. Then $K^{+}\/$ is $W\/$-sound (by assumption), and by Theorem A, $G^{\prime\prime}\/$ has an $EPS\/$-graph $S^{\prime\prime}=E^{\prime\prime}\cup P^{\prime\prime}\/$ such that $K^{+}\subseteq E^{\prime\prime}\/$ and $d_{P^{\prime\prime}}(z)\leq 1\/$ for every $z\in W\/$.

(b1) Suppose $E(P_{4})\cap E(P^{\prime\prime})=\emptyset\/$. Then $P_{4}\subseteq E^{\prime\prime}\/$. Since $\widehat{G}\/$ is a non-trivial block chain, by Lemma 1(ii), $\widehat{G}\/$ contains a $JEPS\/$-graph $\widehat{S}=\widehat{J}\cup\widehat{E}\cup\widehat{P}\/$ such that $d_{\widehat{P}}(w_{3})=0=d_{\widehat{P}}(w^{\prime}_{3})$, and $w_{3},w^{\prime}_{3}$ are the odd vertices of $\hat{J}$; hence $d_{\widehat{J}}(x_{3})=2$ and $d_{\widehat{P}}(x_{3})=0$. Note that by the second part of Lemma 1(ii) we can make sure that $\mbox{min}\{d_{\widehat{P}}(u_{3}),d_{\widehat{P}}(v_{3})\}\leq 1$. In this case, we obtain an $EPS\/$-graph $S=E\cup P\/$ of $G^{+}\/$ by setting $E=(E^{\prime\prime}-P_{4})\cup\widehat{J}\cup\widehat{E}\/$ and $P=P^{\prime\prime}\cup\widehat{P}\/$. Here $d_{P}(x_{3})=0\/$ and $d_{P}(w)\leq 1\/$ for every $w\in W-x_{3}\/$.

(b2) Suppose $E(P_{4})\cap E(P^{\prime\prime})\neq\emptyset$. That is, $V(P_{4})\subseteq V(P^{\prime\prime})$ (so that $E(P_{4})\cap E(E^{\prime\prime})=\emptyset$) and $d_{P^{\prime\prime}}(x_{3})=1\/$. This means that either $a_{2}x_{3}\not\in E(P^{\prime\prime})\/$ or $x_{3}a_{3}\not\in E(P^{\prime\prime})\/$. Suppose $x_{3}a_{3}\not\in E(P^{\prime\prime})\/$ (so that $a_{3}a_{4}\in E(P^{\prime\prime})\/$). In this case, we delete $x_{3}v_{3}$ from $\widehat{G}$ and thus split $\widehat{G}$ into two block chains $\widehat{G_{1}}$ and $\widehat{G_{2}}$ with $x_{3},w_{3}\in\widehat{G_{1}}$ and $v_{3},w^{\prime}_{3}\in\widehat{G_{2}}$. If $\widehat{G_{j}}$ is an edge only, then $\widehat{S_{j}}=\widehat{G_{j}}$. If $\widehat{G_{2}}=\emptyset$, then $\widehat{S_{2}}=\emptyset$. Otherwise by Lemma 1(i) (or by Theorem D if $\widehat{G_{2}}$ is 2-connected), $\widehat{G_{j}}$ has an $EPS$-graph $\widehat{S_{j}}=\widehat{E_{j}}\cup\widehat{P_{j}}$ where $d_{\widehat{P_{1}}}(w_{3})\leq 1\/$, $d_{\widehat{P_{1}}}(x_{3})=1$, $d_{\widehat{P_{2}}}(v_{3})\leq 1$, $d_{\widehat{P_{2}}}(w^{\prime}_{3})\leq 1$, $j=1,2$. Now, if we take $E=\widehat{E_{1}}\cup\widehat{E_{2}}\cup E^{\prime\prime}\/$ and $P=\widehat{P_{1}}\cup\widehat{P_{2}}\cup(P^{\prime\prime}-\{a_{2},a_{3}\})\/$, we have an $EPS\/$-graph $S=E\cup P$ of $G^{+}\/$ with $d_{P}(w)\leq 1\/$ for every $w\in W\/$ (note that $w_{3}a_{2},w^{\prime}_{3}a_{3}\in P^{\prime\prime})$, $x_{3}$ is a pendant vertex in $S$, and it works also if $\widehat{G}$ is a path on at least 4 vertices.

In both cases (b1) and (b2), a required hamiltonian cycle in $(G^{+})^{2}\/$ can be constructed from $S$ (see Observation (*)(i)-(ii)). Note that $G^{+}-x_{4}$ is 2-connected if $K^{+}=yx_{1}x_{4}x_{2}^{*}x_{2}y$ (hence $d_{G}(x_{2})=2$) and if $d_{G}(x_{2}^{*})>2$. Here we have a contradiction to the assumption of this case (1.2)(b).

Now assume that the cycle $K^{+}$ is not $W$-sound. Since $y,x_{1},x_{2}^{*},t_{4}\in K^{+}$ and $|W|=5$, there exists a cycle $K^{*}\subseteq G^{+}$ containing all of $W$ and not $x_{4}$.

(i) Suppose $t_{4}=v_{4}$ or $t_{4}=x_{2}$. In both cases, $K^{*}$ contains $u_{4}=x_{1}$ and $v_{4}=t_{4}$, $v_{4}=x_{2}^{*}$, respectively, but not $x_{4}$. Hence $G^{+}-x_{4}$ is 2-connected, a contradiction with assumptions of case (1.2)(b).

(ii) Suppose $t_{4}=u_{4}$. Because $G^{+}-x_{3}$ has a cutvertex, without loss of generality suppose that $u_{3}\notin K^{+}$ but clearly $u_{3}\in K^{*}$. Hence $u_{3}\notin\{x_{1},x_{2}^{*},u_{4}\}$. We define $x^{**}_{2}$ as $x^{*}_{2}$ with respect to $K^{*}$.

First suppose $x^{**}_{2}=x^{*}_{2}$. By Theorem B, $G^{+}$ has a $[u_{4};x_{1},x_{2}^{*},u_{3}]$-$EPS$-graph $S=E\cup P$. Note that either $x_{4}$ is a pendant vertex in $S$, or else $x_{4}$ is a vertex in $E$. It is clear that $S^{2}$ yields a hamiltonian cycle of $(G^{+})^{2}$ as required (see Observation (*)(i)-(ii)).

If $x^{**}_{2}\neq x^{*}_{2}$, then we proceed analogously as before using $x^{**}_{2}$ instead of $x^{*}_{2}$. Note that $x^{**}_{2}=u_{3}$ or $x^{**}_{2}=u_{4}$ is not an obstacle (we use Theorem C) because of $d_{S}(x_{2})=2$ since $d_{G}(x_{2})=2$ and $x_{2}^{*}$ is also in $K^{*}$ (thus $x_{2}x_{2}^{*}\notin E(S)$); and $x^{**}_{2}=u_{3}=u_{4}$ is not possible in this case.

(2)  Suppose $\{u_{3},v_{3}\}=\{x_{1},x_{2}\}\/$.

Note that, in $G\/$, there exists a cycle containing $x_{3}\/$ and $x_{4}\/$ (and hence also the vertices $u_{3},v_{3},u_{4},v_{4}\/$).

Let $G^{*}=G^{+}-x_{3}\/$ which is homeomorphic to $G\/$ and thus $G^{*}\/$ is $2\/$-connected. Note that there exists a cycle $K^{*}=K^{+}\/$ (see above) in $G^{*}\/$ containing the vertices $y,x_{1},x_{2},x_{4}\/$.

(2.1) Suppose $w\in N(x_{4})-\{x_{1},x_{2}\}\/$ exists; let $x_{2}z_{2}\in E(K^{*})\/$. Note that $d_{G^{*}}(x_{2})=2\/$ if $d_{G}(z_{2})>2\/$.

By Theorem B, there exists an $[x_{1};x_{2},z_{2},w]\/$-$EPS\/$-graph, an $[x_{1};x_{2},z_{2}]\/$-$EPS\/$-graph by Theorem C respectively, if $w=z_{2}$; in both cases we denote $S^{*}=E^{*}\cup P^{*}\subset G^{*}\/$ with $K^{*}\subseteq E^{*}\/$ and $d_{P^{*}}(x_{1})=0\/$. Note that $K^{*}$ is $[x_{1};x_{2},z_{2}]$-maximal if $z_{2}=w$. Set $E=E^{*}\/$ and $P=P^{*}\cup\{x_{1}x_{3}\}\/$; thus $d_{P}(x_{1})=1\/$. Also, $d_{P}(x_{2})+d_{P}(z_{2})\leq 1$ since $d_{P}(z_{2})>0$ implies $d_{P}(x_{2})=0$ since $d_{G^{*}}(x_{2})=2$. Then $S=E\cup P\/$ is an $EPS\/$-graph of $G^{+}\/$ and a hamiltonian cycle in $(G^{+})^{2}\/$ can be constructed (using $S$) which starts with $yx_{1},x_{1}x_{3}\/$, ends with $yx_{2}\/$ and traverses $wx_{4}\/$ even if $w=z_{2}\/$ (see Observation (*)(i)-(ii) for $x_{1}x_{3}$).

(2.2)  Next assume that $\{u_{4},v_{4}\}=\{x_{1},x_{2}\}\/$.

Note that, in this case, $d_{G}(x_{2})>2\/$ since $d_{G}(x_{1})>2\/$ can be assumed and $x_{3},x_{4}\/$ are $2\/$-valent (note that the lemma is trivially true if $G$ is a $4$-cycle).

Consider the graph $G^{\prime}=G-\{x_{3},x_{4}\}\/$.

(a) Suppose $G^{\prime}\/$ is $2\/$-connected. We shall apply Theorem 1 to $G^{\prime}\/$ with $x_{1},x_{2}\/$ in place of $v,w\/$.

(i) Suppose $G^{\prime}\/$ has an $EPS\/$-graph $S^{\prime}=E^{\prime}\cup P^{\prime}\/$ with $d_{P^{\prime}}(x_{i})=0\/$ for $i=1,2\/$. Let $E=E^{\prime}\cup\{yx_{1}x_{4}x_{2}y\}\/$ and $P=P^{\prime}\cup\{x_{1}x_{3}\}\/$; this yields an $EPS\/$-graph $S=E\cup P\/$ of $G^{+}\/$ with $d_{P}(x_{1})=1\/$, $d_{P}(x_{2})=0\/$, $d_{P}(x_{3})=1\/$ and $d_{P}(x_{4})=0\/$. Hence we may construct a hamiltonian cycle in $(G^{+})^{2}\/$ containing the edges $x_{1}x_{3}\/$ and $x_{2}x_{4}\/$ apart from $yx_{1},yx_{2}\/$.

(ii) Suppose $G^{\prime}\/$ has a $JEPS\/$-graph $S^{\prime}=J^{\prime}\cup E^{\prime}\cup P^{\prime}\/$ with $x_{1},x_{2}\/$ being the only odd vertices of $J^{\prime}\/$ and $d_{P^{\prime}}(x_{1})=0=d_{P^{\prime}}(x_{2})\/$. Let $E=E^{\prime}\cup(J^{\prime}\cup\{x_{1}yx_{2}\})\/$ and $P=P^{\prime}\cup\{x_{1}x_{3},x_{2}x_{4}\}\/$. Then $S=E\cup P\/$ is an $EPS\/$-graph of $G^{+}\/$ with $d_{P}(x_{1})=d_{P}(x_{2})=d_{P}(x_{3})=d_{P}(x_{4})=1\/$. Hence a hamiltonian cycle in $(G^{+})^{2}\/$ containing the edges $yx_{1},yx_{2},x_{1}x_{3}\/$ and $x_{2}x_{4}\/$ can be constructed.

(b) Finally assume that $G^{\prime}\/$ is not $2\/$-connected. Then $G^{\prime}\/$ is a non-trivial block chain. By Lemma 1(ii) with $x_{1}=v\/$ and $x_{2}=w\/$, $G^{\prime}\/$ has a $JEPS\/$-graph $S^{\prime}=J^{\prime}\cup E^{\prime}\cup P^{\prime}\/$ with $d_{P^{\prime}}(x_{1})=0=d_{P^{\prime}}(x_{2})\/$. As before, take $E=E^{\prime}\cup(J^{\prime}\cup\{x_{1}yx_{2}\})\/$ and $P=P^{\prime}\cup\{x_{1}x_{3},x_{2}x_{4}\}\/$. Then $S=E\cup P\/$ is an $EPS\/$-graph of $G^{+}\/$ with $d_{P}(x_{1})=d_{P}(x_{2})=d_{P}(x_{3})=d_{P}(x_{4})=1\/$. Hence a hamiltonian cycle in $(G^{+})^{2}\/$ containing the edges $yx_{1},yx_{2},x_{1}x_{3}\/$ and $x_{2}x_{4}\/$ can be constructed.

This completes the proof of the lemma.

Proof of Theorem 2.

Let $G\/$ be a $2\/$-connected $DT\/$-graph and $A=\{x_{1},x_{2},x_{3},x_{4}\}\/$ be a set of four distinct vertices in $G\/$. It is easy to see that the theorem holds if $G\/$ is a cycle. Hence we also apply induction, apart from direct construction at the given graph. However, in general let $G^{+}\/$ be defined as before.

Case (A): $N(x_{i})\subseteq V_{2}(G)\/$, $i=1,2,3,4\/$.

There exists a cycle $K^{+}\/$ in $G^{+}\/$ containing the vertices $y,x_{1},x_{2},x_{4}\/$ (and possibly $x_{3}\/$), assuming that $K^{+}$ is at least as long as any cycle containing $y,x_{1},x_{2},x_{3}\/$. Assume $K^{+}\/$ is $W\/$-sound for $W=\{y,x_{1},x_{2},x_{3},x_{4}\}\/$. By Theorem A, there exists a $W\/$-$EPS\/$-graph $S=E\cup P\/$ in $G^{+}\/$ with $K^{+}\subseteq E\/$ (that is, $d_{P}(w)\leq 1\/$ for every vertex $w\/$ in $W\/$). Moreover $d_{P}(y)=0\/$ (since $y\/$ is $2\/$-valent in $G^{+}\/$ and $K^{+}\subseteq E\/$).

Since $N(x_{i})\subseteq V_{2}(G)\/$ for $i=1,2,3,4\/$, a hamiltonian cycle $C\/$ in $(G^{+})^{2}\/$ can be constructed, and $C\/$ will contain $yx_{1},yx_{2}\/$ and at least one edge of $G\/$ incident to $x_{j}\/$ for $j=3,4\/$. That is, $G^{2}\/$ contains a hamiltonian path as required (see Observation (*)(i)).

Case (B): $N(x_{i})\subseteq V_{2}(G)\/$, $i=1,2,3\/$ and $N(x_{4})\not\subseteq V_{2}(G)\/$; i.e., $d_{G}(x_{4})=2\/$.

Let $K^{+}\/$ be a cycle in $G^{+}\/$ containing $y,x_{1},x_{2},x_{4}\/$ and possibly $x_{3}\/$.

(B)(1) Suppose $x_{3}\/$ is not in $K^{+}\/$ (so, no cycle of $G^{+}\/$ contains $y\/$ and $x_{i}\/$, $i=1,2,3,4\/$).

(a) Suppose $\{u_{4},v_{4}\}\neq\{x_{1},x_{2}\}\/$.

Then we may assume that $u_{4}\not\in\{x_{1},x_{2}\}\/$. Let $G^{\prime}=G^{+}-x_{4}\/$ and let $K^{\prime}\subseteq G^{\prime}\/$ be a cycle containing $y,x_{1},x_{2},x_{3}\/$.

(a1) Suppose $G^{\prime}\/$ is $2\/$-connected.

Set $W^{\prime}=\{y,x_{1},x_{2},x_{3},u_{4}\}\/$ and suppose without loss of generality that $K^{\prime}\/$ is $W^{\prime}\/$-sound (i.e., $u_{4}\in V(K^{\prime})\/$ if $G^{\prime}\/$ has a cycle containing all of $W^{\prime}\/$). By Theorem A, $G^{\prime}\/$ has a $W^{\prime}\/$-$EPS\/$-graph $S^{\prime}=E^{\prime}\cup P^{\prime}\/$ with $K^{\prime}\subseteq E^{\prime}\/$ such that $d_{P^{\prime}}(w)\leq 1\/$ for all $w\in W^{\prime}\/$ with $d_{P^{\prime}}(y)=0\/$. Take $E=E^{\prime}\/$ and $P=P^{\prime}\cup\{u_{4}x_{4}\}\/$. Then $S=E\cup P\/$ is an $EPS\/$-graph of $G^{+}\/$ with $K^{\prime}\subseteq E\/$, $d_{P}(y)=0,d_{P}(x_{4})=1\/$ and $d_{P}(w)\leq 1\/$ for $w\in W^{\prime}-\{u_{4}\}\/$; and $d_{P}(u_{4})\leq 2\/$. A careful examination of this case and Observation (*)(i)-(ii) show that a required hamiltonian cycle in $(G^{+})^{2}\/$ can be constructed (note that $u_{4}x_{4}$ is a pendant edge of $S$).

(a2) Suppose $G^{\prime}\/$ is not $2\/$-connected.

Then $G^{\prime}\/$ is a non-trivial block chain. Let $B_{y}\/$ denote the block in $G^{\prime}\/$ containing $K^{\prime}\/$. Note that $u_{4},v_{4}\/$ belong to different endblocks of $G^{\prime}\/$. Let $z_{4}\in\{u_{4},v_{4}\}\/$ be a vertex in an endblock $B_{1}\/$ of $G^{\prime}\/$ where $B_{1}\neq B_{y}\/$. Further let $\widehat{G}\/$ denote the maximal block chain in $G^{\prime}\/$ containing $B_{1}\/$ but no edges of $B_{y}\/$. Let $c_{0}\in V(B_{y})\cap V(\widehat{G})\/$ be a cutvertex of $G^{\prime}\/$ (which is not a cutvertex of $\widehat{G}\/$).

Now replace $\widehat{G}\/$ in $G^{+}\/$ with a path $P_{2}=z_{4}zc_{0}\/$ of length $2\/$ joining $z_{4}\/$ and $c_{0}\/$ and call the resulting graph $G^{*}\/$; $z\not\in V(G)\/$. In so doing the cycle $K^{+}\/$ is transformed into the cycle $K^{*}\/$ in $G^{*}\/$ containing $P_{2}\cup\{y,x_{1},x_{2},x_{4}\}\/$. Observe that $x_{3}\not\in V(K^{*})\/$; otherwise $K^{*}\/$ could be extended to become a cycle in $G^{+}\/$ containing $y,x_{1},\ldots,x_{4}\/$ contrary to the supposition of this case. Set $W=\{y,x_{1},x_{2},x_{3},x_{4}\}\/$. $K^{*}\/$ is $W\/$-sound in $G^{*}\/$; by Theorem A, $G^{*}\/$ contains a $W\/$-$EPS\/$-graph $S^{*}=E^{*}\cup P^{*}\/$ with $K^{*}\subseteq E^{*}\/$, $d_{P^{*}}(y)=0=d_{P^{*}}(x_{4})=d_{P^{*}}(z_{4})\/$ and $d_{P^{*}}(w)\leq 1\/$ for all $w\in W-\{y,x_{4}\}$.

Let $H=\widehat{G}\cup P_{2}\/$. Then $H\/$ is a $2\/$-connected graph and hence has a $[c_{0};z_{4}]\/$-$EPS\/$-graph $S_{H}=E_{H}\cup P_{H}\/$ with $K_{H}\subseteq E_{H}\/$ where $K_{H}=(K^{+}\cap\widehat{G})\cup P_{2}\/$ (see Theorem D).

By taking $E=((E^{*}\cup E_{H})-(K^{*}\cup K_{H}))\cup K^{+}\/$ and $P=P^{*}\cup P_{H}\/$ we have $S=E\cup P\/$ being a $W\/$-$EPS\/$-graph of $G^{+}\/$ with $K^{+}\subseteq E\/$, $d_{P}(y)=0=d_{P}(x_{4})\/$, $d_{P}(w)\leq 1\/$ for all vertices $w\in W-\{y,x_{4}\}$, $d_{P}(z_{4})\leq 1\/$ and $d_{P}(y_{4})\leq 2\/$ where $y_{4}\in N(x_{4})-z_{4}\/$. Hence a required hamiltonian cycle $H\/$ in $(G^{+})^{2}\/$ can be constructed (as $\{u_{4}x_{4},x_{4}v_{4}\}\subseteq K^{+}\/$); in particular $z_{4}x_{4}\in E(H)\/$ (see Observation (*)(i)).

(b) Suppose $\{u_{4},v_{4}\}=\{x_{1},x_{2}\}\/$.

Let $G^{\prime}=G^{+}-x_{4}\/$ (which is $2\/$-connected since $G\/$ is $2\/$-connected) and let $K^{\prime}\/$ be a cycle in $G^{\prime}\/$ containing $y,x_{1},x_{2},x_{3}\/$. By Theorem C, there exists an $[x_{1};x_{2},x_{3}]\/$-$EPS\/$-graph $S^{\prime}=E^{\prime}\cup P^{\prime}\/$ in $G^{\prime}\/$ with $K^{\prime}\subseteq E^{\prime}\/$, $d_{P^{\prime}}(w)\leq 1\/$ for $w\in\{x_{2},x_{3}\}\/$ and $d_{P^{\prime}}(x_{1})=0\/$. Let $E=E^{\prime}\/$ and $P=P^{\prime}\cup\{x_{1}x_{4}\}\/$. Then we have an $EPS\/$-graph $S=E\cup P\/$ of $G^{+}\/$ with $K^{\prime}\subseteq E\/$ and $d_{P}(x_{i})\leq 1\/$ for $i=1,2,3,\/$ and $x_{4}\/$ is a pendant vertex in $S\/$. Hence we can can construct a hamiltonian cycle in $(G^{+})^{2}\/$ containing the edges $yx_{1},yx_{2},x_{1}x_{4}\/$ and $x_{3}t_{3}\/$ where $t_{3}\in N(x_{3})\/$ since $N(x_{3})\subseteq V_{2}(G)\/$ (see Observation (*)(i)-(ii)).

(B)(2) Suppose also $x_{3}\/$ is in $K^{+}\/$.

Assume without loss of generality that $K^{+}=yx_{1}z_{1}\cdots z_{3}x_{3}w_{3}\cdots u_{4}x_{4}v_{4}\cdots z_{2}x_{2}y\/$.

(a) Suppose $u_{4}\neq x_{3}\/$.

Set $W=\{y,x_{1},x_{2},x_{3},u_{4}\}\/$. Then $W\subseteq K^{+}\/$ and hence $K^{+}\/$ is $W\/$-sound. By Theorem A, there is a $W\/$-$EPS\/$-graph $S=E\cup P\/$ in $G^{+}\/$ such that $K^{+}\subseteq E\/$ and $d_{P}(w)\leq 1\/$ for every $w\in W\/$. Then it is possible to construct in $(G^{+})^{2}\/$ a hamiltonian cycle $C\/$ containing the edges $x_{3}w_{3}\/$ and $u_{4}x_{4}\/$ (recall that $x_{4},w_{3}\/$ are $2\/$-valent vertices in $G\/$) (see Observation (*)(i)).

(b) Suppose $u_{4}=x_{3}\/$.

(i) Suppose $v_{4}\neq x_{2}\/$. We apply Theorem B to $G^{+}\/$ to obtain an $[x_{3};x_{1},x_{2},v_{4}]\/$-$EPS\/$-graph $S=E\cup P\/$ with $K^{+}\subseteq E\/$ and $d_{P}(x_{3})=0\/$, $d_{P}(x_{i})\leq 1\/$ for $i=1,2\/$, and $d_{P}(v_{4})\leq 1\/$. Since $K^{+}\subseteq E\/$ and $x_{4}\in K^{+}\/$, we have $d_{P}(x_{4})=0\/$. We can construct a hamiltonian cycle $C\/$ in $(G^{+})^{2}\/$ whose two edges incident to $x_{i}\/$ are edges of $G\/$ for $i=3\/$ or $i=4\/$, one of which is (without loss of generality) $x_{3}x_{4}\/$ (see Observation (*)(i)).

(ii) Suppose $v_{4}=x_{2}\/$. We operate analogously as in case (i) with an $[x_{3};x_{1},x_{2},y]\/$-$EPS\/$-graph $S\/$ provided $x_{3}\not\in N(x_{1})\/$. However $S^{2}\/$ does not yield a hamiltonian cycle as required if $x_{3}\in N(x_{1})\/$. That is, $d_{G}(x_{3})=2$; $d_{G}(x_{4})=2$, and $N(x_{1})\subseteq V_{2}(G)$ by the assumptions. This is a special case of Lemma 2. This finishes the proof of Case (B).

Case (C): $N(x_{i})\subseteq V_{2}(G)\/$, $i=1,2\/$ and $d_{G}(x_{3})=2=d_{G}(x_{4})\/$.

The proof of this case follows from Lemma 2.

Case (D): $N(x_{1})\subseteq V_{2}(G)\/$ and $N(x_{2})\not\subseteq V_{2}(G)\/$; $d_{G}(x_{2})=2\/$ follows.

**(D)(1)  $N(x_{4})\subseteq V_{2}(G)\/$. **

There is a cycle $K^{+}$ in $G^{+}$ containing $y,x_{1},x_{2},x_{3}$ and also $x_{4}$ if a such a cycle exists. Recall that $x_{3}^{*}=x_{3}$ if $d_{G}(x_{3})>2$ and $x_{3}^{*}=u_{3}=z_{3}\/$ if $d_{G}(x_{3})=2\/$, and $N(x_{2})=\{u_{2},v_{2}\}\/$ and assume that $v_{2}\/$ is in $K^{+}\/$. Let $x_{3}^{-},x_{3}^{+}\/$ denote the predecessor, successor respectively, of $x_{3}\/$ in $K^{+}\/$, where we start the traversal of $K^{+}$ with the edge $yx_{1}$. We also note that $x_{3}^{*}=u_{3}=x_{3}^{-}$ and $v_{3}=x_{3}^{+}$ if $x_{3}\in V_{2}(G)$.

(1.1) Assume that $v_{2}\not\in\{x_{3},x_{4}\}\/$.

(a) $N(x_{3})\subseteq V_{2}(G)\/$.

Let $W=\{y,x_{1},v_{2},x_{3},x_{4}\}\/$. Without loss of generality let $K^{+}\/$ be chosen such that it is $W\/$-sound, since $\{y,x_{1},v_{2},x_{3}\}\subseteq K^{+}\/$ anyway, and possibly $x_{4}\in K^{+}\/$. Let $S=E\cup P\/$ be a $W$-$EPS\/$-graph of $G^{+}\/$ with $K^{+}\subseteq E\/$ (by Theorem A). Observe that if $x_{4}\not\in E\/$, then it is a pendant vertex in $S\/$; also $d_{P}(x_{2})\leq 1\/$ automatically since $N(x_{2})\not\subseteq V_{2}(G)\/$ and $x_{2}\in K^{+}\/$. Now it is easy to construct a required hamiltonian cycle $C\/$ in $S^{2}\/$ having the required properties; we may assume that $x_{3}x_{3}^{+}\in E(C)\/$ and $x_{4}w_{4}\in E(G)\cap E(C)\/$, since $d_{P}(x_{4})\leq 1\/$ and $N(x_{4})\subseteq V_{2}(G)$. This is even true if $x_{1}=x_{3}^{-}\/$ since both $x_{3}\/$ and $x_{3}^{+}\/$ are $2\/$-valent in $G$ in this case (see Observation (*)(i)-(ii)).

(b) $N(x_{3})\not\subseteq V_{2}(G)\/$; $d_{G}(x_{3})=2\/$ follows.

Set $W^{+}=\{y,x_{1},v_{2},x_{3}^{+},x_{4}\}\/$.

(b1) Suppose $x_{4}=x_{3}^{+}\/$. Since $K^{+}\supset\{y,x_{1},v_{2},x_{3},x_{3}^{+},x_{4}\}\/$, by Theorem B, $G^{+}\/$ contains an $[x_{4};y,x_{1},v_{2}]\/$-$EPS\/$-graph $S=E\cup P\/$ with $K^{+}\subseteq E\/$ such that $d_{P}(x_{4})=d_{P}(x_{3})=0$, $d_{P}(v_{2})\leq 1\/$, $d_{P}(x_{1})\leq 1\/$, but also $d_{P}(x_{2})\leq 1$. We obtain a hamiltonian cycle $C\subset S^{2}$ as required with $x_{3}x_{3}^{+},x_{4}w_{4}\in E(G)\cap E(C)$ ($w_{4}\notin V(K^{+})$ may hold, if $d_{S}(x_{4})>2$). This covers also the case $x_{1}x_{3}\in E(K^{+})$.

(b2) Suppose $x_{4}\neq x_{3}^{+}\/$.

(b2.1) Now assume that $K^{+}\/$ is $W^{+}\/$-sound.

(i) Suppose $x_{3}^{+}\neq v_{2}$. Let $S=E\cup P$ be a $W^{+}\/$-$EPS\/$-graph of $G^{+}\/$ with $K^{+}\subseteq E$, by Theorem A. Then $S^{2}\/$ contains a hamiltonian cycle $C\/$ of $(G^{+})^{2}\/$ as required, even if $x_{1}x_{3}\in E(K^{+})$ and $d_{G}(x_{3}^{+})>2\/$. In any case, also here $C\/$ can be constructed from $S\/$ such that $x_{3}x_{3}^{+},x_{4}w_{4}\in E(G)\cap E(C)\/$.

(ii) Suppose $x_{3}^{+}=v_{2}$. Hence $x_{4}\in K^{+}$, otherwise $|V(K^{+})\cap W^{+}|=3$ and $K^{+}$ is not $W^{+}$-sound, a contradiction. Then $G^{+}\/$ contains an $[x_{2};x_{1},x_{3}^{+},x_{4}]$-$EPS$-graph $S=E\cup P$ with $K^{+}\subseteq E$, by Theorem B. Hence we obtain a hamiltonian cycle $C\subset S^{2}$ as required with $x_{3}x_{3}^{+},x_{4}w_{4}\in E(G)\cap E(C)$.

(b2.2) Assume that $K^{+}\/$ is not $W^{+}\/$-sound.

(i) Suppose $|V(K^{+})\cap W^{+}|>3$. Then there exists a cycle $K^{*}\/$ in $G^{+}\/$ containing $y,x_{1},v_{2},x_{3}^{+},x_{4}\/$ but $x_{3}\not\in K^{*}\/$; otherwise, we should have chosen $K^{+}=K^{*}\/$ which is $W^{+}$-sound, a contradiction.

First suppose $x_{2}v_{2}\in E(K^{*})$. By Theorem B (if $x_{3}^{+}\neq v_{2}$), Theorem C (if $x_{3}^{+}=v_{2}$), there is an $[x_{3}^{+};x_{1},v_{2},x_{4}]\/$-$EPS\/$-graph, $[x_{3}^{+};x_{1},x_{4}]\/$-$EPS\/$-graph, respectively, $S=E\cup P\/$ of $G^{+}\/$ with $K^{*}\subseteq E\/$. Note that either $x_{3}\/$ is a vertex in $E\/$, or else it is a pendant vertex in $S\/$. Also take note that $d_{P}(x_{2})\leq 1\/$ and $d_{P}(v_{2})\leq 1\/$. By Observation (*) (i)-(ii), $S^{2}\/$ has a hamiltonian cycle with the required properties.

If $x_{2}v_{2}\notin E(K^{*})$, then $x_{2}u_{2}\in E(K^{*})$ and we proceed analogously as before using $u_{2}$ instead of $v_{2}$. Note that $u_{2}=x_{4}$ is not an obstacle (we use Theorem C) because of $d_{S}(x_{2})=2$ since $d_{G}(x_{2})=2$ and $v_{2}$ is also in $K^{*}$ (thus $x_{2}v_{2}\notin E(S)$); and $v_{2}=x_{3}^{+}$ is not possible in this case.

(ii) Suppose $|V(K^{+})\cap W^{+}|=3$.

Hence $x_{4}\notin K^{+}$ and $v_{2}=x_{3}^{+}$. If $x_{1}x_{3}\notin E(K^{+})$, then we set $W^{*}=\{y,x_{1},x_{3}^{-},x_{3}^{+},x_{4}\}$. If $x_{1}x_{3}\in E(K^{+})$ and $d_{G}(v_{2})=2$, then we set $W^{*}=\{y,x_{1},x_{2},x_{3}^{+},x_{4}\}$. In both cases $K^{+}$ is $W^{*}$-sound. By Theorem A, $G^{+}$ contains a $W^{*}$-$EPS$-graph $S=E\cup P$ with $K^{+}\subseteq E$. Observe that if $x_{4}\not\in E\/$, then it is a pendant vertex in $S\/$. Now it is easy to construct a required hamiltonian cycle $C\/$ in $S^{2}\/$ having the required properties (see Observation (*)(i)-(ii)).

If $x_{1}x_{3}\in E(K^{+})$ and $d_{G}(v_{2})>2$, then we consider $G-x_{3}$.

If $G-x_{3}$ is 2-connected, then we apply induction and get an ${\cal F}_{4}$ $x_{1}x_{2}$-hamiltonian path $P_{1}$ in $(G-x_{3})^{2}$ containing edges $x_{4}w_{4},x_{3}^{+}w_{3}^{+}\in E(G)$. Then

[TABLE]

defines a hamiltonian path in $G^{2}$ as required.

If $G-x_{3}$ is not 2-connected, then $x_{1}$ belongs to one endblock and $x_{3}^{+},x_{2}$ to the other endblock of a non-trivial block chain $G-x_{3}$ because of the degree condition of $x_{3},x_{2}$. Moreover $x_{1}$, $x_{3}^{+}$, and $x_{2}$ are not cutvertices of $G-x_{3}$. Depending on the position of $x_{4}$ in $G-x_{3}$ we construct a hamiltonian path $P$ in $G^{2}$ as in the preceding case applying either induction, or Theorem F, proceeding block after block. Since this procedure is straightforward we do not work out the details.

(1.2) $v_{2}\in\{x_{3},x_{4}\}\/$.

(1.2.1) Assume that $v_{2}=x_{3}\/$.

$G^{+}-x_{2}u_{2}\/$ is a trivial or non-trivial block chain. Let $B_{y}\/$ denote the endblock (in $G^{+}-x_{2}u_{2}\/$) containing the cycle $K^{+}\/$ and let $\widehat{G}=(G^{+}-x_{2}u_{2})-B_{y}\/$.

Suppose first that $\widehat{G}\neq\emptyset$. Hence $\widehat{G}\/$ is a block chain in $G^{+}-x_{2}u_{2}\/$ containing $u_{2}\/$ and $N(u_{2})-x_{2}\/$. Let $t\/$ be the cutvertex of $G^{+}-x_{2}u_{2}\/$ belonging to $B_{y}\/$. That is, $\widehat{G}\cap B_{y}=\{t\}\/$.

(a) Suppose $x_{4}\not\in\widehat{G}-t\/$.

Then $x_{4}\/$ is in $B_{y}\/$. Observe that $G-x_{2}\/$ is a block chain with $\widehat{G}\/$ being an induced subgraph of $G-x_{2}\/$ (note that $d_{G}(x_{2})=2\/$). Since $B_{y}\/$ is $2\/$-connected, it contains a path $P(x_{3},x_{1})\/$ through $x_{4}\/$. It follows that $x_{2},y\not\in P(x_{3},x_{1})\/$. Thus $P(x_{3},x_{1})\subseteq G-x_{2}\/$ with $x_{4}\in P(x_{3},x_{1})\/$. Now, $P(x_{2},x_{1})=x_{2}x_{3}P(x_{3},x_{1})\/$ is a path in $G-\widehat{G}\/$. Thus we may assume that $K^{+}=yx_{2}P(x_{2},x_{1})x_{1}y\subseteq B_{y}\/$ and thus passes through $y,x_{1},x_{2},x_{3}^{*},x_{4}\/$. By Theorem C, $B_{y}\/$ has an $[x_{3}^{*};x_{1},x_{4}]\/$-$EPS\/$-graph $S_{y}=E_{y}\cup P_{y}\/$ with $K^{+}\subseteq E_{y}\/$ and $d_{P_{y}}(x_{2})=0\/$ (note that $d_{B_{y}}(x_{2})=2\/$). Let the same $S_{y}\/$ denote an $[x_{3}^{*};x_{1}]$-$EPS\/$-graph of $B_{y}\/$ if $x_{4}=x_{3}^{*}\/$ (i.e., $x_{3}x_{4}\in E(K^{+})\/$) (see Theorem D).

Since $x_{4}\not\in\widehat{G}-t\/$, by Lemma 1(i) or Theorem D if $\widehat{G}$ is 2-connected, $\widehat{G}$ contains an $EPS\/$-graph $\widehat{S}=\widehat{E}\cup\widehat{P}\/$ with $d_{\widehat{P}}(t)=0\/$ and $d_{\widehat{P}}(u_{2})\leq 1\/$, provided $d_{\widehat{G}}(t)>1$. If $d_{\widehat{G}}(t)=1$, then either $\widehat{S}=\emptyset$ if $\widehat{G}=u_{2}t$ or by Lemma 1(i), $\widehat{G}-t$ contains an $EPS\/$-graph $\widehat{S}=\widehat{E}\cup\widehat{P}\/$ with $d_{\widehat{P}}(t_{1})\leq 1\/$ and $d_{\widehat{P}}(u_{2})\leq 1\/$, where $t_{1}\in N_{\widehat{G}}(t)$.

Since $P_{y}\cap\widehat{P}=\emptyset\/$, by setting $E=E_{y}\cup\widehat{E}\/$ and $P=P_{y}\cup\widehat{P}\cup\{u_{2}x_{2}\}\/$, we obtain an $EPS\/$-graph $S=E\cup P\/$ of $G^{+}\/$ with $K^{+}\subseteq E\/$, $d_{P}(x_{2})=1\/$, $d_{P}(x_{3}^{*})=0\/$, $d_{P}(x_{i})\leq 1\/$ for $i=1,4\/$ and $d_{P}(u_{2})\leq 2\/$.

If $|V(K^{+})|\geq 6$, a required hamiltonian cycle in $S^{2}\/$ can be constructed (note that the cases $x_{4}=t\/$ and $x_{4}\neq t\/$ are treated simultaneously) (see Observation (*)(i)).

If, however, $|V(K^{+})|<6$, i.e., $|V(K^{+})|=5$, then $d_{G}(x_{1})=2\/$ since $N(x_{4})\subseteq V_{2}(G)\/$ and $N(x_{1})\subseteq V_{2}(G)\/$, which in turn implies $d_{G}(x_{4})=d_{G}(x_{3})=2\/$. Hence $G-\{x_{3},x_{4}\}\/$ is a block chain $G^{-}\/$ with $x_{1},x_{2}\/$ being pendant vertices of $G^{-}\/$. It follows that $(G^{-})^{2}\/$ has a hamiltonian path $HP^{-}\/$ starting with $x_{1}w_{1}\in E(G)\/$ and ending with $x_{2}u_{2}\in E(G)\/$. Clearly,

[TABLE]

defines a required hamiltonian path in $G^{2}\/$.

(b) Suppose $x_{4}\in\widehat{G}-t\/$.

In this case, we note that in $B_{y}$, the cycle $K^{+}$ can be assumed to traverse $y,x_{1},t,x_{3},x_{2}$ in this order; it also contains $x_{3}^{*}$ if $x_{3}$ is $2$-valent. As for $t\in V(K^{+})\/$, see the preceding observation at the beginning of (a), with $t$ assuming the role of $x_{4}$.

Suppose $t\neq x_{1}$. By Theorem C, $B_{y}$ has an $[x_{3}^{*};x_{1},t]$-$EPS$-graph $S_{y}=E_{y}\cup P_{y}$ with $K^{+}\subseteq E_{y}\/$ if $x_{3}^{*}\neq t$; by Theorem D, $B_{y}$ has an $[x_{3}^{*};x_{1}]$-$EPS$-graph $S_{y}=E_{y}\cup P_{y}$ with $K^{+}\subseteq E_{y}$ if $x_{3}^{*}=t$; and $d_{P_{y}}(x_{2})=0$ since $d_{G}(x_{2})=2$.

Suppose $t=x_{1}$. We let $S_{y}=E_{y}\cup P_{y}$ be an $[x_{1};x_{3}^{*}]$-$EPS$-graph in $B_{y}$ with $K^{+}\subseteq E_{y}$ by Theorem D. Note that we set $x_{3}^{*}=x_{2}$ if $x_{1}x_{3}\in E(K^{+})$.

(b1) Assume that $x_{4}\/$ is a cutvertex in $\widehat{G}\/$.

(i) Consider the case $x_{4}\/$ is not incident to a bridge of $\widehat{G}$. Let $\widehat{G_{1}}\/$ and $\widehat{G_{2}}\/$ be defined by $\widehat{G}=\widehat{G_{1}}\cup\widehat{G_{2}}\/$ with $t,x_{4}\in V(\widehat{G_{1}})\/$, $x_{4},u_{2}\in V(\widehat{G_{2}})\/$ and $\widehat{G_{1}}\cap\widehat{G_{2}}=\{x_{4}\}\/$.

By Lemma 1(i) or Theorem D, $\widehat{G_{i}}\/$ has an $EPS\/$-graph $\widehat{S_{i}}=\widehat{E_{i}}\cup\widehat{P_{i}}\/$ with $d_{\widehat{P_{i}}}(x_{4})=0\/$ for $i=1,2\/$, $d_{\widehat{P_{1}}}(t)\leq 1\/$ and $d_{\widehat{P_{2}}}(u_{2})\leq 1\/$.

By taking $E=E_{y}\cup\widehat{E_{1}}\cup\widehat{E_{2}}\/$, $P=P_{y}\cup\widehat{P_{1}}\cup\widehat{P_{2}}\/$, we have an $EPS\/$-graph $S=E\cup P\/$ of $G^{+}\/$ with $d_{P}(x_{2})=0=d_{P}(x_{4})\/$, and $d_{P}(x_{1})\leq 1\/$ and $d_{P}(x_{3}^{*})\leq 1\/$; $d_{P}(t)\leq 2\/$ by construction, provided $t\neq x_{1}\/$. Moreover, if $t=x_{3}^{*}$, we have $d_{P_{y}}(x_{3}^{*})=0\/$ and hence $d_{P}(x_{3}^{*})\leq 1\/$ because of $d_{\widehat{P_{1}}}(x_{3}^{*})\leq 1\/$; and $d_{P}(x_{1})\leq 1\/$. Also if $t=x_{1}\/$, we have $d_{P_{y}}(x_{1})=0\/$ and hence $d_{P}(x_{1})\leq 1\/$ because of $d_{\widehat{P_{1}}}(t)\leq 1\/$; and $d_{P}(x_{3}^{*})\leq 1.$ Hence a required hamiltonian cycle in $S^{2}\/$ can be constructed (the various construction details are straightforward and are thus omitted).

(ii) Now suppose $x_{4}\/$ is incident to a bridge $f\/$ of $\widehat{G}$ and $|V(K^{+})|>4\/$. In this case, we delete $f\/$ and thus split $\widehat{G}\/$ into two block chains $\widehat{G_{1}}\/$ and $\widehat{G_{2}}\/$ with $t\in\widehat{G_{1}}\/$, $u_{2}\in\widehat{G_{2}}\/$ and $x_{4}\/$ is either in $\widehat{G_{1}}$ or in $\widehat{G_{2}}$. By Lemma 1(i) or Theorem D, $\widehat{G_{i}}$ has an $EPS$-graph $\widehat{S_{i}}=\widehat{E_{i}}\cup\widehat{P_{i}}$ with $d_{\widehat{P_{1}}}(t)\leq 1$, $d_{\widehat{P_{2}}}(u_{2})\leq 1$ and $d_{\widehat{P_{i}}}(x_{4})\leq 1$ for some $i\in\{1,2\}$. Note that $\widehat{S_{i}}=G_{2}$ if $G_{i}=K_{2}$; or $\widehat{S_{i}}=\emptyset$ if $G_{i}=t$ or $G_{i}=u_{2}$. Proceeding similarly to case (i) let $E=E_{y}\cup\widehat{E_{1}}\cup\widehat{E_{2}}$ and $P=P_{y}\cup\widehat{P_{1}}\cup(\widehat{P_{2}}\cup\{u_{2}x_{2}\})\/$. Then we have an $EPS\/$-graph $S=E\cup P\/$ of $G^{+}$.

Because of the choice of $S_{y}\/$ in the cases $t\notin\{x_{1},x_{3}^{*}\}\/$, $t=x_{3}^{*}$, and $t=x_{1}$, we have in any case, $d_{P}(x_{1})\leq 1\/$, $d_{P}(x_{2})=1\/$, $d_{P}(x_{3}^{*})\leq 1\/$ and $x_{4}\/$ is either a pendant vertex in $S\/$ or $d_{P}(x_{4})=0\/$ (which occurs when $d_{G}(x_{4})>2\/$).

By a similar argument as in case (i), we conclude that in all cases $S^{2}\/$ contains a hamiltonian cycle with the required properties unless $d_{P}(x_{2})=1$, $d_{P}(x_{3}^{*})=1$ and $x_{3}^{*}=x_{3}=t$. In this case there exists a cycle containing $y,x_{1},x_{2},x_{3},x_{4}$, a contradiction to the choice of $K^{+}$.

(iii) Now suppose $x_{4}\/$ is incident to a bridge $f\/$ of $\widehat{G}$ and $|V(K^{+})|=4\/$. It follows that $t=x_{1}$ and therefore $G^{\prime}=G-x_{3}\/$ is a non-trivial block chain containing $f$ as a bridge. Hence $(G-x_{3})^{2}\/$ contains a hamiltonian path $HP^{\prime}\/$ starting with an edge $x_{1}w_{1}\in E(G)\/$ and ending with $u_{2}x_{2}\in E(G)\/$ and containing $f\/$. It follows that

[TABLE]

yields a hamiltonian path in $G^{2}\/$ as required if $f\neq x_{1}x_{4}$. However, if $f=x_{1}x_{4}$, then we set

[TABLE]

(b2) Hence assume that $x_{4}\/$ is not a cutvertex in $\widehat{G}\/$.

Suppose first that $x_{4}\/$ is contained in a $2\/$-connected block $B\/$ in $\widehat{G}\/$. Further, let $c_{1},c_{2}\/$ be two vertices in $B\/$ which are also cutvertices of $\widehat{G}\/$ if $B\/$ is not an endblock of $\widehat{G}\/$. If, however, $B\/$ is an endblock of $\widehat{G}\/$, then let $c_{1}\/$ be the unique cutvertex of $\widehat{G}\/$ in $B\/$, and let $c_{2}\in\{t,u_{2}\}\/$ depending on which of the endblocks of $\widehat{G}\/$ is $B\/$. If $x_{4}\neq c_{2}$, we apply Theorem C to $B\/$ to obtain an $[x_{4};c_{1},c_{2}]\/$-$EPS\/$-graph of $B\/$; if $x_{4}=c_{2}$ (which means $x_{4}=u_{2}$), then we apply Theorem D to $B\/$ to obtain an $[x_{4};c_{1}]\/$-$EPS\/$-graph of $B\/$. In both cases by using Lemma 1(i), extend these $EPS\/$-graphs to an $EPS\/$-graph $\widehat{S}=\widehat{E}\cup\widehat{P}\/$ of $\widehat{G}\/$ with $d_{\widehat{P}}(t)\leq 1\/$, $d_{\widehat{P}}(u_{2})\leq 1\/$, and $d_{\widehat{P}}(x_{4})=0\/$.

Setting $E=E_{y}\cup\widehat{E}\/$ and $P=P_{y}\cup\widehat{P}\/$, we obtain an $EPS\/$-graph $S=E\cup P\/$ of $G^{+}\/$ with $d_{P}(x_{2})=0=d_{P}(x_{4})\/$, $d_{P}(x_{3}^{*})\leq 1\/$ and $d_{P}(x_{1})\leq 1\/$.

Hence assume that $x_{4}\/$ is not contained in a $2$-connected block. That is, $x_{4}\/$ is a pendant vertex in $\widehat{G}\/$. In this case, $x_{4}=u_{2}\/$. We apply Lemma 1(i) to obtain an $EPS$-graph $\widehat{S}=\widehat{E}\cup\widehat{P}$ of $\widehat{G}\/$ with $d_{\widehat{P}}(t)\leq 1$, and $d_{\widehat{P}}(x_{4})\leq 1\/$ if $\widehat{G}\neq x_{4}t$. If $\widehat{G}=x_{4}t$, then $\widehat{S}=\widehat{G}$. Setting $E=E_{y}\cup\widehat{E}$ and $P=P_{y}\cup\widehat{P}$, we obtain an $EPS$-graph $S=E\cup P\/$ of $G^{+}\/$ with $d_{P}(x_{2})=0\/$ and $d_{P}(x_{3}^{*})\leq 1\/$, $d_{P}(x_{1})\leq 1$, $d_{P}(x_{4})=1\/$ and $x_{4}\/$ is a pendant vertex in $S$.

In any of these cases, $S^{2}\/$ contains a hamiltonian cycle $C\/$ with the required properties (note that $x_{3}x_{2}\in E(C)\/$ because $d_{E}(x_{2})=d_{G^{+}}(x_{2})-1=2\/$); see Observation(*)(i)-(ii).

Finally if $\widehat{G}=\emptyset$, we find $S_{y}$ as in Case (1.2.1)(a) and construct a hamiltonian cycle as required using $S_{y}$ only.

(1.2.2) Assume that $v_{2}=x_{4}\/$.

Recall that the cycle $K^{+}$ in $G^{+}$ contains $y,x_{1},x_{2},x_{3},v_{2}$. Therefore

[TABLE]

Consider the graph $G^{\prime}=G^{+}-x_{2}u_{2}$.

Case (a)  $G^{\prime}\/$ is $2\/$-connected.

Suppose $x_{3},x_{4}\/$ are adjacent in $K^{+}\/$. Then apply Theorem D to obtain an $[x_{4};x_{1}]\/$-$EPS\/$-graph $S=E\cup P\/$ of $G^{\prime}\/$ with $K^{+}\subseteq E\/$. Suppose $x_{3},x_{4}\/$ are not adjacent in $K^{+}\/$. Then apply Theorem C to obtain an $[x_{1};x_{3}^{*},x_{4}]\/$-$EPS\/$-graph $S=E\cup P\/$ of $G^{\prime}\/$ with $K^{+}\subseteq E\/$. In either case, a required hamiltonian cycle in $S^{2}\/$ can be constructed (setting $x_{3}^{*}=x_{3}^{+}=w_{3}$ if $x_{1}x_{3}\in E(K^{+})\/$).

Case (b)  $G^{\prime}\/$ is not $2\/$-connected.

Then $G^{\prime}\/$ is a non-trivial block chain. As before, let $B_{y}\/$ denote the endblock in $G^{\prime}\/$ containing $y\/$ (and hence containing the cycle $K^{+}\/$). Set $\widehat{G}=G^{\prime}-B_{y}\/$ which is a trivial or non-trivial block chain; $\widehat{G}\neq\emptyset$ in any case. It follows that $B_{y}\cap\widehat{G}=\{t\}\/$ and $t\/$ is a cutvertex of $G^{\prime}\/$. By Theorem D or Lemma 1(i), $\widehat{G}\/$ has an $EPS\/$-graph $\widehat{S}=\widehat{E}\cup\widehat{P}$ with $d_{\widehat{P}}(t)\leq 1$ if $\widehat{G}\neq u_{2}t$. If $\widehat{G}=u_{2}t$, then $\widehat{S}=\widehat{G}$.

(i) Suppose $t=x_{4}\/$.

Then $G^{\prime\prime}=G^{+}-x_{2}x_{4}\/$ is $2\/$-connected. Replace in $K^{+}\/$ the edge $x_{4}x_{2}\/$ with a path $P(x_{4},x_{2})\/$ in $\widehat{G}\cup\{u_{2}x_{2}\}\/$ to obtain the cycle $K^{\prime\prime}\/$. Since $\{y,x_{1},x_{3},x_{4},u_{2},x_{2}\}\subseteq V(K^{\prime\prime})\/$, we may apply Theorem B to obtain an $[x_{4};x_{1},u_{2},x_{2}]\/$-$EPS\/$-graph $S^{\prime\prime}=E^{\prime\prime}\cup P^{\prime\prime}\subseteq G^{\prime\prime}\/$ if $x_{3}\/$ and $x_{4}\/$ are adjacent in $K^{\prime\prime}\/$, or to obtain an $[x_{1};x_{3}^{*},x_{4},u_{2}]\/$-$EPS\/$-graph $S^{\prime\prime}=E^{\prime\prime}\cup P^{\prime\prime}\subseteq G^{\prime\prime}$ if $x_{3}\/$ and $x_{4}\/$ are not adjacent in $K^{\prime\prime}\/$ (setting $x_{3}^{*}=x_{3}^{+}=w_{3}$ if $x_{1}x_{3}\in E(K^{\prime\prime})\/$). In both cases, $K^{\prime\prime}\subseteq E^{\prime\prime}\/$. A required hamiltonian cycle in $(S^{\prime\prime})^{2}$ can be constructed (since the situation is similar to Case (a) above); see Observation (*)(i).

(ii) Suppose $t=x_{3}\/$.

We apply Theorem C to $B_{y}\/$ to obtain an $[x_{3};x_{1},x_{4}]\/$-$EPS\/$-graph $S_{y}=E_{y}\cup P_{y}\/$ of $G^{\prime}$ with $K^{+}\subseteq E_{y}\/$. Note that $N(x_{3})\subseteq V_{2}(G)\/$ in this case.

(iii) Suppose $t=x_{1}\/$ and $x_{1}x_{3}x_{4}\not\subseteq K^{+}\/$.

We set $x_{3}^{*}=x_{3}^{+}=w_{3}$, if $x_{1}x_{3}\in E(K^{+})$. We apply Theorem C to $B_{y}\/$ again to obtain an $[x_{1};x_{3}^{*},x_{4}]\/$-$EPS\/$-graph $S_{y}=E_{y}\cup P_{y}\/$ of $G^{\prime}$ with $K^{+}\subseteq E_{y}\/$.

In the cases (ii) and (iii), we let $E=\widehat{E}\cup E_{y}\/$, $P=\widehat{P}\cup P_{y}\/$ and obtain an $EPS\/$-graph $S=E\cup P\/$ of $G^{+}-x_{2}u_{2}$ with $K^{+}\subseteq E\/$, $d_{P}(x_{2})=0\/$ and $d_{P}(w)\leq 1\/$ for $w\in\{x_{1},x_{3}^{*},x_{4}\}\/$. Hence a required hamiltonian cycle in $S^{2}\/$ can be constructed; see Observation (*)(i).

(iv) Suppose $t\not\in\{x_{1},x_{3},x_{4}\}\/$.

We set $x_{3}^{*}=x_{3}^{+}=w_{3}$ if $x_{1}x_{3}\in E(K^{+})$. Note that $d_{B_{y}}(x_{2})=2$. Let $W=\{y,x_{1},x_{3}^{*},x_{4},t\}$; $W-\{t\}\subseteq V(K^{+})$.

First suppose that $|W|=5$. If $K^{+}$ is not $W$-sound, then there is a cycle $K^{\prime}\subseteq B_{y}$ containing all vertices of $W$, in which case we apply Theorem B to $B_{y}\/$ to obtain an $[x_{3}^{*};x_{1},x_{4},t]\/$-$EPS\/$-graph $S_{y}=E_{y}\cup P_{y}\/$ with $K^{\prime}\subseteq E_{y}\/$. Note that, if $x_{3}\notin K^{\prime}$, then either $x_{3}$ is a pendant vertex in $S_{y}$ or $d_{P_{y}}(x_{3})=0$ and $x_{3}^{*}x_{3}\in E(E_{y})$. If $K^{+}\/$ is $W\/$-sound, then we set $K^{\prime}=K^{+}$ and apply Theorem A to $B_{y}\/$ to obtain a $W\/$-$EPS\/$-graph $S_{y}=E_{y}\cup P_{y}\/$ with $K^{\prime}\subseteq E_{y}\/$.

Suppose $|W|=4$. Hence $x_{3}^{*}\neq x_{3}$.

If $x_{3}^{*}=x_{4}\neq t$, then $N_{G}(x_{3})=\{x_{1},x_{4}\}$ and $B_{y}-x_{3}$ is 2-connected: for, there are two internally disjoint paths from $t$ to $K^{+}$, and the endvertices of these paths in $K^{+}$ are $x_{1}$ and $x_{4}$ since $x_{3},x_{2}\in V_{2}(G)$. Thus $B_{y}-x_{3}$ contains a cycle $K^{\prime}$ containing $y,x_{1},x_{2},x_{4},t$. Hence we apply Theorem B to $B_{y}-x_{3}$ to obtain an $[x_{1};x_{2},x_{4},t]$-$EPS$-graph $S^{\prime}_{y}=E^{\prime}_{y}\cup P^{\prime}_{y}\/$ with $K^{\prime}\subseteq E^{\prime}_{y}\/$. Let $E_{y}=E^{\prime}_{y}$ and $P_{y}=P^{\prime}_{y}\cup\{x_{1}x_{3}\}$. Thus we have an $EPS$-graph $S_{y}=E_{y}\cup P_{y}\/$ of $B_{y}$ with $K^{\prime}\subseteq E_{y}\/$. Moreover $d_{P_{y}}(x_{1})=1$, $d_{P_{y}}(x_{2})=0$, $d_{P_{y}}(x_{3})=1$, $d_{P_{y}}(x_{4})\leq 1$, $d_{P_{y}}(t)\leq 1$, and $x_{3}$ is a pendant vertex in $S_{y}$.

If $x_{3}^{*}=t\notin\{x_{1},x_{3},x_{4}\}$, then we set $K^{\prime}=K^{+}$ and apply Theorem C to $B_{y}$ to obtain an $[x_{3}^{*};x_{1},x_{4}]$-$EPS$-graph $S_{y}=E_{y}\cup P_{y}$ with $K^{\prime}\subseteq E_{y}$.

In all cases, we let $E=\widehat{E}\cup E_{y}\/$, $P=\widehat{P}\cup P_{y}\/$ and obtain an $EPS\/$-graph $S=E\cup P\/$ of $G^{+}-x_{2}u_{2}\/$ with $K^{\prime}\subseteq E\/$, $d_{P}(x_{2})=0\/$ and $d_{P}(w)\leq 1\/$ for $w\in\{x_{1},x_{3}^{*},x_{4}\}\/$ (even if $x_{1}x_{3}\in E(K^{+})\/$ or $x_{3}^{*}=x_{3}^{+}=t\neq x_{3}\/$). Hence a required hamiltonian cycle in $S^{2}\/$ can be constructed; see Observation (*)(i)-(ii).

(v) Suppose $t=x_{1}\/$ and $x_{1}x_{3}x_{4}\subseteq K^{+}\/$.

Note that $K^{+}=yx_{1}x_{3}x_{4}x_{2}y\/$. Let $G_{3}=B_{y}-\{y,x_{2}\}\/$; it is $2\/$-connected if $d_{G}(x_{4})>2\/$, or else it is a path $x_{1}x_{3}x_{4}\/$. We have $G-x_{2}x_{4}=G_{3}\cup(\widehat{G}\cup\{u_{2}x_{2}\})\/$ with $t=G_{3}\cap\widehat{G}\/$. Consequently,

[TABLE]

By Corollary 1(ii), $\widetilde{G}^{2}\/$ has a hamiltonian path $P_{1,2}\/$ starting with $x_{1}w_{1}\in E(G)\/$ and ending with $u_{2}x_{2}\/$. If $G_{3}\/$ is $2\/$-connected, then $G_{3}-x_{3}\/$ is a (trivial or non-trivial) block chain and thus $(G_{3}-x_{3})^{2}\/$ has a hamiltonian path $P_{4,1}\/$ starting in $x_{4}=v_{2}$ and ending with an edge $s_{1}x_{1}\in E(G)$ (using Theorem F(ii) if $G_{3}-x_{3}$ is 2-connected, Corollary 1(ii) if $G_{3}-x_{3}$ is a non-trivial block chain, and $P_{4,1}=G_{3}-x_{3}$ if $G_{3}-x_{3}=x_{1}x_{4}$). Set

[TABLE]

if $G_{3}=x_{1}x_{3}x_{4}\/$; and

[TABLE]

if $G_{3}\/$ is $2$-connected. In both cases, $P(x_{1},x_{2})\/$ is a ${\cal F}_{4}\/$ $x_{1}x_{2}\/$-hamiltonian path in $G^{2}\/$. This finishes the proof of Case (D)(1).

Since the case $N(x_{3})\subseteq V_{2}(G)\/$ is analogous to the Case (D)(1), we are left with the following case.

(D)(2) $N(x_{i})\not\subseteq V_{2}(G)\/$ for $i=3,4\/$.

However, the proof of this case follows from Lemma 2. This finishes the proof of Case (D).

Case (E): $N(x_{1})\not\subseteq V_{2}(G)\/$ and $N(x_{2})\not\subseteq V_{2}(G)\/$.

Then $d_{G}(x_{1})=2=d_{G}(x_{2})\/$.

Let $K^{+}\/$ be a cycle containing the vertices $y,u_{1},u_{2},x_{3}\/$ and possibly $x_{4}\/$ where we assume that

[TABLE]

**(E)(1) ** Suppose $x_{4}\/$ is not in any cycle containing $y\/$ and $x_{3}\/$.

(1.1)  $d_{G}(x_{3})>2,\ d_{G}(x_{4})>2\/$.

(a) Suppose $x_{3}\not\in\{u_{1},u_{2}\}\/$.

Set $W=\{y,u_{1},u_{2},x_{3},x_{4}\}$. By supposition, $K^{+}$ is $W$-sound. By Theorem A, we have an $EPS$-graph $S=E\cup P$ with $K^{+}\subseteq E$ and $d_{P}(w)\leq 1\/$ for every $w\in W\/$. In this case a required hamiltonian cycle $C\/$ in $S^{2}\/$ can be constructed (taking note that $x_{1},x_{2}\/$ are $3\/$-valent in $G^{+}\/$, and that $x_{i}x_{4}\in E(P)\/$, $i\in\{1,2\}\/$, does not constitute an obstruction in the construction of $C\/$).

(b) Suppose $x_{3}=u_{1}\/$.

Note that if $x_{4}\not\in N(x_{1})\cup N(x_{2})\/$, then we are back to case (a) with $x_{3}\/$ and $x_{4}\/$ changing roles. Hence we have $x_{4}\in\{v_{1},v_{2}\}$. Also, $x_{4}=v_{1}=v_{2}\/$ cannot hold; otherwise, $d_{G}(x_{4})>2\/$ and $x_{i}\in N(x_{4})\/$, $d_{G}(x_{i})=2\/$, $i=1,2\/$ imply the existence of an $x_{4}q_{3}\/$-path $P(x_{4},q_{3})\subset G\/$ with $q_{3}\in V(K^{+})\/$ and $(P(x_{4},q_{3})-q_{3})\cap K^{+}=\emptyset\/$, yielding in turn a cycle containing $y,x_{3},x_{4}\/$ contradicting E(1). By the same token, $x_{3}=u_{1}=u_{2}\/$ cannot hold.

(b1) $x_{4}=v_{2}\/.$

Consider $G^{-}=G-\{x_{1}u_{1},x_{2}u_{2}\}\/$.

Note that $x_{3},x_{4}\/$ belong to different components of $G^{-}\/$; otherwise there is a path $P_{0}\/$ in $G^{-}\/$ joining $x_{3}\/$ and $x_{4}\/$ implying that $C_{0}=P_{0}x_{4}x_{2}yx_{1}x_{3}\/$ is a cycle in $G^{+}\/$ with $y,x_{3},x_{4}\in V(C_{0})\/$, a contradiction to the supposition. Since $G\/$ is $2$-connected, $G^{-}$ contains precisely two components $G_{3}^{-}\neq K_{1}$ and $G_{4}^{-}$ containing $x_{3},x_{4}$, respectively. Clearly $x_{2}\in V(G_{4}^{-})$. We also have $x_{1}\in V(G_{4}^{-})$ because $P_{0}$ as above does not exist.

Observe that $G_{4}^{-}$ and $G_{3}^{-}$ are (trivial or non-trivial) block chains in which $x_{1},x_{2}\in V(G_{4}^{-})$ and $x_{3},u_{2}\in V(G_{3}^{-})$ are not cutvertices. Thus $G^{+}-\{x_{1}u_{1},x_{2}u_{2}\}\/$ is a disconnected graph with two components $G_{3}=G_{3}^{-}\/$ (which contains $x_{3}=u_{1}\/$ and $u_{2}\/$) and $G_{4}\/$ (which contains $y,x_{1},v_{1},x_{4}=v_{2}\/$ and $x_{2}\/$).

Note that in $G_{4}\/$, there is a cycle $C^{+}$ containing $y,x_{1},v_{1},x_{4},x_{2}$, implying that $G_{4}\/$ is $2\/$-connected, whereas $G_{3}\/$ is a block chain. By Theorem D, let $S_{4}=E_{4}\cup P_{4}$ be a $[v_{1};x_{4}]$-$EPS$-graph in $G_{4}$ with $C^{+}\subseteq E_{4}$, $d_{P_{4}}(v_{1})=0=d_{P_{4}}(x_{1})=d_{P_{4}}(x_{2})$ and $d_{P_{4}}(x_{4})\leq 1$. By Lemma 1(i) or Theorem D (respectively depending on whether $G_{3}$ has a cutvertex or $G_{3}$ is $2$-connected), there is an $EPS\/$-graph $S_{3}=E_{3}\cup P_{3}\/$ in $G_{3}\/$ such that $d_{P_{3}}(x_{3})=0\/$ and $d_{P_{3}}(u_{2})\leq 1\/$. Taking $E=E_{3}\cup E_{4}\/$ and $P=P_{3}\cup\{x_{1}x_{3}\}\cup P_{4}$, we have an $EPS$-graph $S=E\cup P$ of $G^{+}\/$ with $C^{+}\subseteq E\/$ and $d_{P}(v_{1})=0=d_{P}(x_{2})\/$, $d_{P}(x_{1})=1=d_{P}(x_{3})$ and $d_{P}(x_{4})\leq 1\/$.

Note that in this case, since $d_{G}(x_{3}),\ d_{G}(x_{4})>2\/$, $d_{P}(x_{2})=0$, $d_{P}(x_{3})=1\/$ and $d_{P}(x_{4})\leq 1\/$, it is straightforward that one can obtain a required hamiltonian cycle of $(G^{+})^{2}\/$.

(b2) $x_{4}=v_{1}$.

Let $G^{\prime}=G^{+}-x_{1}x_{3}\/$, and we may assume that a cycle $K^{\prime}=yx_{1}x_{4}\cdots v_{2}x_{2}y\subseteq G^{\prime}$ exists. Note that in $G$ we have two internally disjoint paths $x_{1}x_{3}\cdots u_{2}x_{2}$ and $x_{1}x_{4}\cdots v_{2}x_{2}$. This is in line with the notation of $K^{+}$ above.

(b2.1) Suppose $G^{\prime}\/$ is $2\/$-connected.

Take $W=\{y,x_{3},x_{4},v_{2},x_{2}\}\/$. Then $K^{\prime}\/$ is $W$-sound in $G^{\prime}\/$ since $v_{2}\neq x_{4}\/$ (see the observation in (b)). Let $S=E\cup P\/$ be a $W\/$-$EPS\/$-graph of $G^{\prime}\/$ (and hence a $W\/$-$EPS\/$-graph of $G^{+}\/$) with $K^{\prime}\subseteq E\/$ and $d_{P}(w)\leq 1\/$ for every $w\in W\/$. Since $d_{G}(x_{3})>2$, $d_{G}(x_{4})>2$, a hamiltonian cycle in $S^{2}\/$ can be constructed containing $x_{1}y,x_{2}y$ and $x_{i}z_{i}$ where $z_{i}\in N_{G}(x_{i})$, $i=3,4$.

(b2.2) Suppose $G^{\prime}\/$ is not $2\/$-connected.

By symmetry, $G^{+}-x_{1}x_{4}\/$ is also not $2\/$-connected. Then $G^{\prime}\/$ is a block chain with endblocks $B_{3},B^{\prime}\/$, with $x_{3}\in B_{3}\/$ and $K^{\prime}\subset B^{\prime}$ and $x_{1}\/$ and $x_{3}\/$ are not cutvertices of $G^{\prime}\/$. Furthermore, let $c\/$ denote the cutvertex of $G^{\prime}\/$ which belongs to $B^{\prime}\/$; $c\neq x_{4}$ (otherwise, $G^{+}\/$ contains a cycle through $y,x_{3},x_{4}\/$).

Set $G_{0}=G^{\prime}-B^{\prime}\/$. Note that $x_{3},c\/$ are vertices in $G_{0}\/$ and are not cutvertices of $G_{0}\/$. By Lemma 1(i) or Theorem D (depending on whether $G_{0}$ has a cutvertex or not), $G_{0}$ contains an $EPS\/$-graph $S_{0}=E_{0}\cup P_{0}\/$ with $d_{P_{0}}(c)\leq 1\/$ and $d_{P_{0}}(x_{3})=0\/$ ($B_{3}\/$ is $2\/$-connected because $d_{B_{3}}(x_{3})>1\/$).

(i) Suppose $c\not\in\{v_{2},x_{2}\}\/$. Let $W^{\prime}=\{y,x_{4},c,v_{2},x_{2}\}\/$. $B^{\prime}\supseteq K^{\prime}\supset(W^{\prime}-c)\/$ in any case. So, $K^{\prime}\/$ is $W^{\prime}$-sound, or there is a cycle $K^{\prime\prime}\supset W^{\prime}\/$ with $B^{\prime}\supseteq K^{\prime\prime}\/$, in which case $K^{\prime\prime}\/$ is $W^{\prime}\/$-sound in $B^{\prime}\/$.

(ii) Now suppose $c=x_{2}$. Set $W^{\prime}=\{y,x_{1},x_{2},v_{2},x_{4}\}\/$ and observe that $K^{\prime}\/$ is $W^{\prime}$-sound in $B^{\prime}\/$ again.

In both cases, we obtain by Theorem A an $EPS$-graph $S^{\prime}=E^{\prime}\cup P^{\prime}$ of $B^{\prime}$ with $K^{\prime}\subseteq E^{\prime}$ or $K^{\prime\prime}\subseteq E^{\prime}$, and $d_{P^{\prime}}(w)\leq 1\/$ for every $w\in W^{\prime}$. Note that if $c\notin\{v_{2},x_{2}\}$, $c\notin K^{\prime}$ and $x_{2}v_{2}\notin E(K^{\prime\prime})$, or if $c=x_{2}$, then $d_{P^{\prime}}(x_{2})=0$ because $d_{B^{\prime}}(x_{2})=2$.

Set $E=E_{0}\cup E^{\prime}$, $P=P_{0}\cup P^{\prime}$ to obtain an $EPS$-graph $S=E\cup P$ of $G^{+}$ with $K^{*}\subseteq E$ where $K^{*}\in\{K^{\prime},K^{\prime\prime}\}$, $d_{P}(x_{3})=0$, $d_{P}(z)\leq 1$ for every $z\in\{y,x_{4},v_{2},x_{2}\}\/$, and $d_{P}(c)\leq 2\/$ if $c\not\in\{v_{2},x_{2}\}\/$, and $d_{P}(c)\leq 1\/$ if $c=x_{2}\/$. Also, $d_{P}(x_{1})=0\/$ since $x_{1}x_{3}\not\in E(S)$. Since $N(x_{i})\subseteq V_{2}(G)$, $i=3,4$ and $d_{P}(x_{3})=0$, $d_{P}(x_{4})\leq 1$, a hamiltonian cycle in $S^{2}$ containing the edges incident to $y$ and containing edges $x_{i}z_{i}$, can be constructed, where $z_{i}\in N_{G}(x_{i})$, $i=3,4$. Observe that $d_{P}(v_{2})=d_{P}(x_{2})=1$ does not create any obstacle.

(iii) Suppose $c=v_{2}\/$. In this case, by Theorem C we take in $B^{\prime}\/$ a $[v_{2};x_{2},x_{4}]\/$-$EPS\/$-graph and proceed as in case (i).

(1.2) $d_{G}(x_{3})>2,d_{G}(x_{4})=2$.

Let $K^{\prime}$ be a cycle in $G^{+}\/$ containing the vertices $y,x_{1},w_{1},u_{4},x_{4},v_{4},w_{2},x_{2}\/$ in this order where $w_{i}\in\{u_{i},v_{i}\}\/$, $i=1,2\/$.

(a) $x_{4}\notin\{w_{1},w_{2}\}\/$

(a1) Suppose $v_{4}\not\in N(x_{2})\/$. Note that in this case $|V(K^{\prime})|>6\/$.

Set $W=\{y,w_{1},w_{2},x_{3},v_{4}\}\/$ and observe that $|W|=5$ and $|K^{\prime}\cap W|\geq 4$.

Suppose $K^{\prime}$ is $W$-sound in $G^{+}$. Then by Theorem A, $G^{+}$ has a $W$-$EPS$-graph $S=E\cup P$ with $K^{\prime}\subseteq E$ and $d_{P}(y)=0=d_{P}(x_{4})$. Moreover, for $i=1,2$, we have $d_{P}(x_{i})\leq 1$ since $d_{G}(x_{i})=2$. Hence we can construct a hamiltonian cycle in $S^{2}$ having the required properties.

Now we assume that $K^{\prime}$ is not $W$-sound. Then there is a cycle $K^{*}$ in $G^{+}$ containing all of $W$ but not containing $x_{4}$. Consider $G^{\prime}=G^{+}-x_{4}$.

(i) Suppose $G^{\prime}$ is $2$-connected. By Theorem B, $G^{\prime}\/$ has a $[v_{4};x_{3},w_{1},w_{2}]\/$-$EPS\/$-graph $S^{\prime}=E^{\prime}\cup P^{\prime}\/$ with $K^{*}\subseteq E^{\prime}\/$. Set $E=E^{\prime}\/$ and $P=P^{\prime}\cup\{v_{4}x_{4}\}\/$ to obtain an $EPS\/$-graph of $G^{+}\/$ with $K^{*}\subseteq E\/$ and $v_{4}x_{4}\/$ is a pendant edge in $S\/$. Hence a hamiltonian cycle in $S^{2}\/$ with the required properties can be constructed. For $i=1,2$, note that if $w_{i}x_{i}\notin E(K^{*})$, then $d_{P}(x_{i})=0$ since $d_{G}(x_{i})=2$ and $w_{i}\in K^{*}$. Observe also that $v_{4}x_{i}\in E(K^{*})$ and $x_{3}x_{i}\in E(K^{*})$ do not constitute any obstacle in this case.

(ii) Suppose $G^{\prime}\/$ is not $2\/$-connected. Let $B_{y}\/$ be the endblock in (the non-trivial block chain) $G^{\prime}\/$ containing $K^{*}\/$, and let $t_{4}\/$ be the cutvertex of $G^{\prime}\/$ belonging to $B_{y}\/$. Set $\widehat{G}=(G^{\prime}-B_{y})\cup\{u_{4}x_{4}\}\/$. Note that $\widehat{G}\/$ is a non-trivial block chain and $\widehat{G}=(G^{+}-B_{y})-x_{4}v_{4}\/$.

Set $W^{*}=\{y,w_{1},w_{2},x_{3},t_{4}\}\/$ and observe that $x_{3}\not\in\{w_{1},w_{2}\}\/$; otherwise, $G^{+}\/$ has a cycle containing $y,x_{3},x_{4}\/$ (contradicting E(1)). In any case, $\widehat{G}\/$ has an $EPS\/$-graph $\widehat{S}=\widehat{E}\cup\widehat{P}\/$ with $d_{\widehat{P}}(t_{4})\leq 1\/$ and $d_{\widehat{P}}(x_{4})=1\/$ by Lemma 1(i).

Now if $t_{4}\in\{w_{1},w_{2},x_{3}\}$, let $S_{y}=E_{y}\cup P_{y}$ be a $[t_{4};r_{4},s_{4},y]$-$EPS$-graph of $B_{y}$ with $K^{*}\subseteq E_{y}$ where $\{r_{4},s_{4},t_{4}\}=\{w_{1},w_{2},x_{3}\}$, by Theorem B.

If, however, $t_{4}\notin\{w_{1},w_{2},x_{3}\}$, we may assume without loss of generality that $K^{*}$ is $W^{*}$-sound (since $|W^{*}|=5$ and $K^{*}\supset W^{*}-t_{4}$). Consequently, let in this case $S_{y}=E_{y}\cup P_{y}\/$ be a $W^{*}$-$EPS\/$-graph of $B_{y}\/$ with $K^{*}\subseteq E_{y}\/$.

In all cases, let an $EPS$-graph $S=E\cup P$ of $G^{+}$ be defined by $E=E_{y}\cup\widehat{E}$, $P=P_{y}\cup\widehat{P}\/$. We have $K^{*}\subseteq E\/$ and note that $d_{P}(w)\leq 1\/$ for every $w\in W^{*}-t_{4}\/$, and $d_{P}(t_{4})\leq 2\/$ but $d_{P}(t_{4})\leq 1\/$ if $t_{4}\in\{w_{1},w_{2},x_{3}\}\/$. It is now straightforward to see that in each of the cases in question, $S^{2}\/$ contains a hamiltonian cycle as required (see the argument at the end of case (i); moreover, $t_{4}x_{i}\in E(K^{*})$ does not constitute an obstacle, $i=1,2$). This finishes case (a1).

Since the case $u_{4}\notin N(x_{1})$ can be treated analogously, we are led to the following case.

(a2) $u_{4}=w_{1}$ and $v_{4}=w_{2}$. Then $|V(K^{\prime})|=6$. In view of case (a1), we may assume that any cycle in $G^{+}$ containing $y,x_{1},x_{2},u_{4},x_{4},v_{4}\/$ has length $6\/$.

Suppose $H=G^{+}-x_{1}u_{4}$ is $2$-connected. Then $H$ has a cycle $C$ containing the edges $u_{4}x_{4},x_{4}v_{4},yx_{1},yx_{2},x_{1}w^{\prime}_{1}\/$ (where $w^{\prime}_{1}\neq u_{4}$). But this means that $|V(C)|>6\/$ (because at least $2\/$ more edges are required to form the cycle $C\/$), a contradiction.

Thus $H$ is not 2-connected, and let $B_{y}\/$ and $B_{4}\/$ denote the endblocks of $H\/$ containing $y\/$ and $x_{4}\/$, respectively.

Suppose $x_{2}$ is not a cutvertex of $H$. Since $\kappa(B_{y})\geq 2$, it follows that $\{x_{2},u_{2},v_{2}\}\subset V(B_{y})$. Now, we have a path $P=P(v_{2},u_{2})$ in $B_{y}\/$ with $x_{2}\notin V(P)$. Since $d_{G}(x_{4})=2\/$, $x_{4}\not\in V(P)\/$; otherwise $u_{4}\in V(P)\/$ as well and hence $x_{4}u_{4}\in E(B_{y}\cap B_{4})$ which is impossible. Thus we obtain for $\{r_{2},w_{2}\}=\{u_{2},v_{2}\}\/$ a cycle

[TABLE]

in $G^{+}\/$ containing $V(K^{\prime})$ and $|V(K^{*})|>6\/$, contradicting the assumption at the beginning of this case. Thus $x_{2}\/$ is a cutvertex of $H\/$.

Observing that $d_{G^{+}}(x_{2})=3\/$ and $\kappa(B_{y})\geq 2\/$, we conclude $d_{B_{y}}(x_{2})=2\/$ and thus $x_{2}w_{2}\in E(H)-E(B_{y})\/$ is the other block of $H$ containing the cutvertex $x_{2}$. It now follows that $B_{y}\cap B_{4}=\emptyset$ since $x_{2}w_{2}\not\in E(B_{4})\/$. Without loss of generality $w_{2}=v_{2}\/$; hence $u_{2}x_{2}\in E(B_{y})\/$.

It now follows that $H-B_{y}\/$ is either a path of length $3\/$, or it is a block chain with $B_{4}\/$ being $2\/$-connected and $x_{2}v_{2}\/$ being a block.

(a2.1) Suppose $x_{3}\in V(B_{y})$. Let $K_{y}$ be a cycle in $B_{y}$ containing $y,x_{1},x_{2},x_{3}$ where we may assume that

[TABLE]

Note that $x_{3}=w^{\prime}_{1}=w^{\prime}_{2}$ is impossible because of $d_{G}(x_{3})>2$. If $x_{3}\neq w^{\prime}_{1}$ and $x_{3}\neq w^{\prime}_{2}$, then $B_{y}$ has an $[x_{3};y,w^{\prime}_{1},w^{\prime}_{2}]$-$EPS$-graph $S_{y}=E_{y}\cup P_{y}$ with $K_{y}\subseteq E_{y}$ by Theorem B. If $x_{3}=w^{\prime}_{1}$ or $x_{3}=w^{\prime}_{2}$, then $B_{y}$ has an $[x_{3};y,w^{\prime}_{2}]$-$EPS$-graph or an $[x_{3};y,w^{\prime}_{1}]$-$EPS$-graph $S_{y}=E_{y}\cup P_{y}$ with $K_{y}\subseteq E_{y}$ by Theorem C, respectively. Likewise, if $d_{G}(u_{4})>2\/$, then $B_{4}\/$ has a $[u_{4};v_{4}]\/$-$EPS\/$-graph $S_{4}=E_{4}\cup P_{4}\/$ with $K^{(4)}\subseteq E_{4}\/$ where $K^{(4)}\/$ is a cycle in $B_{4}\/$ containing $u_{4},x_{4},v_{4}$, by Theorem D. If, however, $B_{4}$ is a bridge of $H$, then the path $P_{4}=u_{4}x_{4}v_{4}\/$ has the only $EPS\/$-graph $S_{4}=E_{4}\cup P_{4}\/$ with $E_{4}=\emptyset\/$.

Setting $E=E_{y}\cup E_{4}$ and $P=P_{y}\cup P_{4}\cup\{x_{1}u_{4}\}$, we have an $EPS$-graph $S=E\cup P$ of $G^{+}$ with $d_{P}(x_{1})=1$, $d_{P}(x_{2})=d_{P}(x_{3})=d_{P}(y)=0$, $d_{P}(w^{\prime}_{1})\leq 1\/$, $d_{P}(w^{\prime}_{2})\leq 1\/$, $d_{P}(x_{4})\in\{0,2\}\/$, $d_{P}(u_{4})\leq 2\/$ and $d_{P}(v_{4})\leq 1\/$. However, $d_{P}(x_{4})=2\/$ implies $d_{P}(v_{4})=1\/$ and thus $x_{4}v_{4}\/$ is a pendant edge. Hence a hamiltonian cycle in $S^{2}\/$ with the required properties can be constructed.

(a2.2) Suppose $x_{3}\in V(B_{4})$; thus $B_{4}$ is $2$-connected. Let $K_{y}$ be a cycle in $B_{y}$ containing $y,x_{1},x_{2}$ where we may assume that

[TABLE]

Note that if $w^{\prime}_{1}=w^{\prime}_{2}$, then $d_{B_{y}}(w^{\prime}_{1})=2$. If $w^{\prime}_{1}\neq w^{\prime}_{2}$, then $B_{y}$ has an $[x_{1};y,w^{\prime}_{1},w^{\prime}_{2}]$-$EPS$-graph $S_{y}=E_{y}\cup P_{y}$ with $K_{y}\subseteq E_{y}$ by Theorem B. If $w^{\prime}_{1}=w^{\prime}_{2}$, then $B_{y}$ has an $[x_{1};y,w^{\prime}_{1}]$-$EPS$-graph $S_{y}=E_{y}\cup P_{y}$ with $K_{y}\subseteq E_{y}$ by Theorem C. Likewise, $B_{4}\/$ has a $[u_{4};x_{3},v_{4}]\/$-$EPS\/$-graph $S_{4}=E_{4}\cup P_{4}\/$ with $K^{(4)}\subseteq E_{4}$ where $K^{(4)}$ is a cycle in $B_{4}$ containing $x_{3},u_{4},x_{4},v_{4}$. Setting $E=E_{y}\cup E_{4}$ and $P=P_{y}\cup P_{4}\cup\{x_{1}u_{4}\}$, we have an $EPS\/$-graph $S=E\cup P\/$ of $G^{+}\/$ and $S^{2}$ contains a hamiltonian cycle as required.

(b) $x_{4}\in\{w_{1},w_{2}\}\/$ but $w_{1}\neq w_{2}\/$.

Without loss of generality assume $x_{4}=w_{1}$ and hence $x_{1}x_{4}\in E(G)$ (the case $x_{4}=w_{2}$, $w_{1}\neq w_{2}$, can be solved by a symmetrical argument). Note that $x_{3}=u_{1}=u_{2}$ cannot hold (see the argument in case (1.1)(b)).

(b1) Suppose $v_{4}\not\in N(x_{2})\/$; i.e., $v_{4}\neq w_{2}\/$. Let $K^{\prime}\/$ be a cycle in $G^{+}\/$ containing $y,x_{1},x_{4},v_{4},w_{2},x_{2}\/$ in this order and let $W=\{y,x_{4},v_{4},w_{2},x_{3}\}$. Then $K^{\prime}$ is $W$-sound because of the supposition at the beginning of (E)(1). By Theorem A, $G^{+}$ has a $W$-$EPS$-graph $S=E\cup P$ with $K^{\prime}\subseteq E\/$ and hence a hamiltonian cycle in $S^{2}\/$ with the required properties can be constructed.

(b2) Suppose $v_{4}=w_{2}$. Assume first that $d_{G}(v_{4})=2$. Let $K^{\prime}$ be the cycle $yx_{1}x_{4}w_{2}x_{2}y\/$ and let $W=\{y,x_{1},x_{2},x_{3},x_{4}\}\/$. Then $K^{\prime}\/$ is $W\/$-sound. By Theorem A, $G^{+}\/$ has a $W\/$-$EPS\/$-graph with $K^{\prime}\subseteq E\/$.

Now assume that $d_{G}(v_{4})>2\/$. Let $z\in N(v_{4})-\{x_{4},x_{2}\}\/$. There is a path $P(v_{4},x_{1})\/$ in $G\/$ from $v_{4}\/$ to $x_{1}\/$ via the vertex $z\/$ since $G\/$ is $2\/$-connected; $x_{2}\not\in P(v_{4},x_{1})\/$ since $d_{G}(x_{2})=2\/$. Now $K^{*}=P(v_{4},x_{1})x_{1}yx_{2}v_{4}\/$ is a cycle in $G^{+}\/$ containing $N(x_{4})\/$ but not $x_{4}\/$ itself. Hence $G^{\prime\prime}=G^{+}-x_{4}\/$ is $2\/$-connected.

We may assume that $K^{+}$ is also a cycle in $G^{\prime\prime}$ containing $y,x_{1},u_{1},x_{3},u_{2},x_{2}$ in this order. If $x_{3}\neq u_{1}$ and $x_{3}\neq u_{2}$, then by Theorem C, $G^{\prime\prime}$ has an $[x_{3};u_{1},u_{2}]$-$EPS$-graph $S^{\prime\prime}=E^{\prime\prime}\cup P^{\prime\prime}$ with $K^{+}\subseteq E^{\prime\prime}$. If $x_{3}=u_{1}$ or $x_{3}=u_{2}$, then by Theorem D, $G^{\prime\prime}$ has an $[x_{3};u_{2}]$-$EPS$-graph or an $[x_{3};u_{1}]$-$EPS$-graph $S^{\prime\prime}=E^{\prime\prime}\cup P^{\prime\prime}$ with $K^{+}\subseteq E^{\prime\prime}$, respectively.

Set $E=E^{\prime\prime}$ and $P=P^{\prime\prime}\cup\{x_{1}x_{4}\}\/$. Then $S=E\cup P\/$ is an $EPS\/$-graph of $G^{+}\/$ such that $d_{P}(x_{1})=1\/$, $d_{P}(x_{3})=0=d_{P}(y)\/$ and $d_{P}(w)\leq 1\/$ for $w\in\{x_{2},u_{1},u_{2}\}\/$ and $x_{1}x_{4}\/$ is a pendant edge in $S\/$. In either case, a hamiltonian cycle in $S^{2}\/$ with the required properties can be constructed.

(c) $N(x_{4})=\{x_{1},x_{2}\}\/$.

Clearly $G^{\prime\prime}=G^{+}-x_{4}\/$ is $2\/$-connected. Let $K^{\prime\prime}\/$ be a cycle in $G^{\prime\prime}\/$ containing $y,x_{1},x_{2},x_{3}\/$, and let $u_{1}\in V(K^{\prime\prime})\cap N_{G}(x_{1})\/$. Without loss of generality, $u_{1}\neq x_{3}\/$: for $d_{G}(x_{3})>2\/$ implies $\{x_{1},x_{2}\}\not\subset N(x_{3})\/$.

Then $G^{\prime\prime}\/$ has an $[x_{3};u_{1}]$-$EPS$-graph $S^{\prime\prime}=E^{\prime\prime}\cup P^{\prime\prime}$ with $K^{\prime\prime}\subseteq E^{\prime\prime}$. Set $E=E^{\prime\prime}\/$ and $P=P^{\prime\prime}\cup\{x_{1}x_{4}\}\/$. Then $S=E\cup P\/$ is an $EPS\/$-graph of $G^{+}\/$ with $d_{P}(y)=0=d_{P}(x_{3})=d_{P}(x_{2})\/$ and $x_{1}x_{4}\/$ being a pendant edge in $S\/$. Hence a hamiltonian cycle in $S^{2}\/$ with the required properties can be constructed.

(1.3) $d_{G}(x_{3})=2,\ d_{G}(x_{4})=2\/$.

Recall that $x_{3},x_{4}$ are not on the same cycle containing $y,x_{1},x_{2}\/$. For each $i=3,4\/$, let $l_{i}\/$ denote the length of a longest cycle in $G^{+}\/$ containing $y,x_{1},x_{2},x_{i}\/$.

(a) Suppose $l_{3}\geq 7$ or $l_{4}\geq 7$; without loss generality assume that $l_{3}\geq 7$. Recall that

[TABLE]

Then either $u_{1}\not\in\{u_{3},x_{3}\}\/$ or $u_{2}\not\in\{v_{3},x_{3}\}\/$. Without loss of generality, assume that $u_{1}\not\in\{u_{3},x_{3}\}\/$.

(a1) Assume that $G^{\prime}=G^{+}-x_{4}\/$ is $2\/$-connected.

Set $W=\{y,u_{1},u_{2},u_{3},q_{4}\}$, where $q_{4}\in\{u_{4},v_{4}\}$. Note that $|\{y,u_{1},u_{2},u_{3}\}|=4$.

Suppose $q_{4}$ exists such that $|W|=4$, say for $q_{4}=u_{4}$. Then $u_{4}\in\{u_{1},u_{2},u_{3}\}$ and $G^{\prime}$ has a $[u_{4};w_{1},w_{2}]$-$EPS$-graph $S^{\prime}=E^{\prime}\cup P^{\prime}$ with $K^{+}\subseteq E^{\prime}$, where $\{u_{4},w_{1},w_{2}\}=\{u_{1},u_{2},u_{3}\}$, by Theorem C.

Now suppose that $|W|=5$ and $K^{+}$ is $W$-sound in $G^{\prime}$ for some choice of $q_{4}$, say for $q_{4}=u_{4}$. Then by Theorem A there is a $W$-$EPS$-graph $S^{\prime}=E^{\prime}\cup P^{\prime}$ of $G^{\prime}$ with $K^{+}\subseteq E^{\prime}$.

In both cases, taking $E=E^{\prime}$ and $P=P^{\prime}\cup\{x_{4}u_{4}\}$, we have an $EPS$-graph $S=E\cup P$ of $G^{+}$ such that $d_{P}(w)\leq 1$ for all $w\in W-\{u_{4}\}$, $d_{P}(x_{4})=1$ and $d_{P}(u_{4})\leq 2$. Hence a required hamiltonian cycle in $S^{2}$ can be constructed; it can be made to contain $x_{4}u_{4}$ and $u_{3}x_{3}$.

Hence we assume that $|W|=5$ and $K^{+}$ is not $W$-sound in $G^{\prime}$ for any choice of $q_{4}\in\{u_{4},v_{4}\}$. Then there is another cycle $K^{\prime}$ in $G^{\prime}$ such that $V(K^{\prime})\supseteq W$. We may assume that $q_{4}=u_{4}$ and $x_{3}\notin K^{\prime}$. Then by Theorem B, $G^{\prime}$ contains a $[u_{3};u_{1},u_{2},u_{4}]$-$EPS$-graph $S^{\prime}=E^{\prime}\cup P^{\prime}$ with $K^{\prime}\subseteq E^{\prime}$. Taking $E=E^{\prime}$ and $P=P^{\prime}\cup\{x_{4}u_{4}\}$, we have an $EPS$-graph $S=E\cup P$ of $G^{+}$. Note that $x_{4}$ is a pendant vertex in $S$ and either $x_{3}$ is a vertex in $E$, or else it is a pendant vertex in $S$. Hence a required hamiltonian cycle in $S^{2}$ can be constructed. For $i=1,2$, also note that if $u_{i}x_{i}\notin E(K^{\prime})$, then $d_{P}(x_{i})=0$ since $d_{G}(x_{i})=2$ and $u_{i}\in K^{\prime}$. Observe also that $u_{3}x_{i}\in E(K^{\prime})$ and $u_{4}x_{i}\in E(K^{\prime})$ do not constitute any obstacle in this case.

(a2) Assume that $G^{\prime}=G^{+}-x_{4}\/$ is not $2\/$-connected.

In view of case (a1), we may assume, by symmetry, that $G^{+}-x_{3}\/$ is also not $2\/$-connected.

Let $K^{(i)}$ denote a cycle containing $y,x_{1},x_{2},x_{i}\/$ where $i\in\{3,4\}\/$. Let $B_{i}\/$ be the block of $G^{+}-x_{i}\/$ with $K^{(7-i)}\subset B_{i}$. Let $G_{i},\ G_{i}^{\prime}\/$ denote the block chains in $G^{+}-x_{i}-B_{i}\/$ (possibly $G_{i}=\emptyset\/$ or $G_{i}^{\prime}=\emptyset\/$) which contain $\{u_{i},c_{i}\}\/$ and $\{v_{i},c_{i}^{\prime}\}$ respectively, where $c_{i},c_{i}^{\prime}$ denote the cutvertices of $G^{+}-x_{i}$ belonging to $B_{i}$, provided $G_{i}\neq\emptyset$, $G_{i}^{\prime}\neq\emptyset\/$. If $G_{i}=\emptyset\/$, then $u_{i}=c_{i}\/$ and is not a cutvertex, and likewise $v_{i}=c_{i}^{\prime}\/$ if $G_{i}^{\prime}=\emptyset\/$.

We observe that $K^{(7-i)}\/$ is edge-disjoint from $G_{i}\cup G_{i}^{\prime}\/$, $i=3,4\/$ and that $G_{3}\cup G_{3}^{\prime}\/$ and $G_{4}\cup G_{4}^{\prime}\/$ are edge-disjoint (since every block of $G_{i}\cup G_{i}^{\prime}$ contains an edge of $K^{(i)}$). Finally, if $C_{y}$ (in $G^{+}$) is a cycle containing $y$, then $E(C_{y}\cap(G_{i}\cup G_{i}^{\prime}))=\emptyset\/$ for at least one $i\in\{3,4\}\/$; otherwise, $C_{y}\supset\{x_{3},x_{4}\}\/$, contrary to (E)(1). Without loss of generality $C_{y}$ is one of the cycles $K^{(3)}$, and we may also assume that $K^{(3)}=K^{+}$ (see the beginning of (a)).

Set $W=\{y,u_{1},u_{2},u_{3},x_{4}\}$. The definition of $W$ together with the last sentences of the preceding paragraph ensure that $|W|=5$ and $K^{(3)}=K^{+}$ is $W$-sound in $G^{+}$.

Set $G_{0}=G_{4}\cup G_{4}^{\prime}\cup\{u_{4}x_{4}v_{4}\}\/$; it is a block chain.

(a2.1) Suppose $G_{0}\/$ is a path with $3\leq l(G_{0})\leq 4\/$.

Then by Theorem A, $G^{+}\/$ has a $W\/$-$EPS\/$-graph $S=E\cup P\/$ with $K^{(3)}\subseteq E\/$ and $d_{P}(x_{4})\leq 1\/$. If $d_{P}(x_{4})=0\/$, then $x_{4}\/$ is in $E\/$, and one of its neighbors is $2\/$-valent because $l(G_{0})\geq 3\/$. If $d_{p}(x_{4})=1\/$, then $x_{4}\/$ is a pendant vertex in $S\/$. In either case, a required hamiltonian cycle in $(G^{+})^{2}\/$ can be constructed.

(a2.2) Suppose $G_{0}\/$ is a path with $l(G_{0})\geq 5\/$, or $G_{0}\/$ is a block chain having non-trivial blocks.

Replace $G_{0}\/$ in $G^{+}\/$ by a path $P_{4}=c_{4}u_{4}x_{4}v_{4}c_{4}^{\prime}\/$ to obtain the graph $G^{*}$ (note that $|E(G_{0})|\geq 5$). Observe that the cycle $K^{(3)}\/$ in $G^{*}\/$ passes through the vertices $y,x_{1},x_{2},x_{3}\/$. Then as in case (a2.1), $G^{*}\/$ has a $W\/$-$EPS\/$-graph $S^{*}=E^{*}\cup P^{*}\/$ with $d_{P^{*}}(x_{4})=0\/$ or $d_{P^{*}}(x_{4})=1\/$.

(i) If $d_{P^{*}}(x_{4})=0\/$, then $P_{4}\subset E^{*}\/$. Since $G_{0}\/$ is a block chain, by Lemma 1(ii), $G_{0}\/$ contains a $JEPS\/$-graph $S_{0}=J_{0}\cup E_{0}\cup P_{0}\/$ such that $d_{P_{0}}(c_{4})=0=d_{P_{0}}(c_{4}^{\prime})\/$. Moreover, in constructing $S_{0}\/$ by proceeding block by block, one can achieve $d_{P_{0}}(u_{4})\leq 1\/$, $d_{P_{0}}(v_{4})\leq 1\/$. In this case, we obtain a $W\/$-$EPS\/$-graph $S=E\cup P\/$ of $G^{+}\/$ by setting $E=(E^{*}-P_{4})\cup J_{0}\cup E_{0}\/$ and $P=P^{*}\cup P_{0}\/$. Here $d_{P}(x_{4})=0\/$, $d_{P}(u_{4})\leq 1,d_{P}(v_{4})\leq 1,d_{P}(c_{4})\leq 2,d_{P}(c_{4}^{\prime})\leq 2\/$ and a required hamiltonian cycle in $(G^{+})^{2}\/$ can be constructed.

(ii) If $d_{P^{*}}(x_{4})=1\/$, then $V(P_{4})\subseteq V(P^{*})\/$. Hence either $u_{4}x_{4}\not\in E(P^{*})\/$ or $v_{4}x_{4}\not\in E(P^{*})\/$. Suppose $v_{4}x_{4}\not\in E(P^{*})\/$ (so that $u_{4}x_{4}\in E(P^{*})\/$). By Lemma 1(i), $G_{4}\cup\{u_{4}x_{4}\}\/$ (respectively $G_{4}^{\prime}\/$) has an $EPS\/$-graph $S^{(4)}=E^{(4)}\cup P^{(4)}\/$ (respectively $S^{\prime(4)}=E^{\prime(4)}\cup P^{\prime(4)}\/$) such that $d_{P^{(4)}}(c_{4})\leq 1,\ d_{P^{(4)}}(u_{4})\leq 2,\ d_{P^{(4)}}(x_{4})=1\/$ with $u_{4}x_{4}\/$ being a pendant edge in $S^{(4)}\/$ and $d_{P^{\prime(4)}}(c_{4}^{\prime})\leq 1,\ d_{P^{\prime(4)}}(v_{4})\leq 1\/$. Now, if we take $E=E^{*}\cup E^{(4)}\cup E^{\prime(4)}\/$ and $P=(P^{*}-\{u_{4},v_{4}\})\cup P^{(4)}\cup P^{\prime(4)}\/$, we have an $EPS\/$-graph $S=E\cup P\/$ of $G^{+}\/$ with $d_{P}(w)\leq 1\/$ for every $w\in W\/$ from which a required hamiltonian cycle in $(G^{+})^{2}\/$ can be constructed (take note that $c_{4}u_{4},v_{4}c_{4}^{\prime}\in E(P^{*})\/$ resulting in $d_{P}(c_{4})\leq 2\/$ and $d_{P}(c_{4}^{\prime})\leq 2\/$; and $d_{P}(x_{i})\leq 1\/$ is guaranteed by the assumption $d_{G}(x_{i})=2\/$, $i=1,2\/$).

In view of case (1.3)(a) solved, we may assume from now on that $l_{3}\leq l_{4}$ and hence we are left with the following case.

(b) Suppose $4\leq l_{3}\leq l_{4}\leq 6\/$.

(b1) Suppose $l_{3}=6$.

(b1.1) Suppose $u_{3}=u_{4}=u_{1}$ and $v_{3}=v_{4}=v_{2}$. Set $G^{*}=G-x_{4}\/$.

$\kappa(G^{*})=2\/$ since $N(x_{4})=N(x_{3})\/$. By induction, $G^{*}\/$ has the ${\cal F}_{4}$-property; that is, there exists an $x_{1}x_{2}\/$-hamiltonian path $P(x_{1},x_{2})\/$ in $(G^{*})^{2}\/$ containing different edges $x_{3}z_{3},u_{4}z_{4}\in E(G^{*})\/$. We may write

[TABLE]

where $\{s,t\}=\{u_{4},z_{4}\}.$ Then

[TABLE]

is a required hamiltonian path in $G^{2}$; it contains $x_{3}z_{3}$ because $P(x_{1},x_{2})$ does.

(b1.2) Suppose $u_{3}=u_{1},\ v_{3}=v_{2}\/$ and $u_{4}=v_{1},\ v_{4}=u_{2}\/$.

Consider $G^{-}=G^{+}-\{x_{1}v_{1},x_{2}u_{2}\}\/$. If there is a path $P(s,t)\/$ from $s\in\{v_{1},u_{2}\}\/$ to $t\in\{u_{1},v_{2}\}\/$ in $G^{-}\/$, then either $l_{3}>6\/$ or $l_{4}>6\/$, or $G^{+}\/$ has a cycle containing both $x_{3}\/$ and $x_{4}\/$. Thus $x_{3}\/$ and $x_{4}\/$ belong to different components of $G^{-}\/$. Let $G_{i}\/$ denote the component of $G^{-}\/$ containing the vertices $u_{i},x_{i},v_{i}\/$, $i\in\{3,4\}\/$. We reach the same conclusion when considering $G^{+}-\{x_{1}u_{1},x_{2}v_{2}\}\/$ instead of $G^{-}\/$. Since $N_{G}(x_{1})\not\subseteq V_{2}(G)\/$, $N_{G}(x_{2})\not\subseteq V_{2}(G)\/$, we may assume without loss of generality that $d(v_{1})>2\/$ or $d(u_{2})>2$ (otherwise, $x_{3}$ and $x_{4}$ switch their roles) and hence both $v_{1},u_{2}\/$ are not $2\/$-valent (otherwise, $v_{1}\/$ or $u_{2}\/$ would be a cutvertex of $G\/$). It follows that $G_{4}\/$ is $2\/$-connected. Likewise, $G_{3}$ is also 2-connected.

There is a cycle $C^{(4)}$ in $G_{4}$ containing $u_{4},x_{4},v_{4}$ and there is a cycle $C^{(3)}$ in $G_{3}$ containing $y,x_{1},x_{2},u_{3},x_{3},v_{3}$. By Theorem D, $G_{i}$ has a $[u_{i};v_{i}]$-$EPS$-graph $S_{i}=E_{i}\cup P_{i}$ with $C^{(i)}\subseteq E_{i}$, $i=3,4$. Note that $d_{P_{3}}(z)=0$ for $z\in\{y,x_{1},x_{2},x_{3}\}$.

Now set $E=E_{3}\cup E_{4}$ and $P=P_{3}\cup P_{4}\cup\{x_{1}v_{1}\}$. Then $S=E\cup P\/$ is an $EPS\/$-graph of $G^{+}\/$ with $C^{(3)}\cup C^{(4)}\subseteq E\/$ and a required hamiltonian cycle in $(G^{+})^{2}\/$ containing $x_{4}v_{1},x_{3}v_{2}\/$ can be constructed.

(b1.3) Suppose $u_{3}=u_{1}=u_{4},v_{3}=v_{2}$ and $v_{4}=u_{2}$ (the case $u_{3}=u_{1}$, $u_{4}=v_{1}$ and $v_{2}=v_{3}=v_{4}$ is symmetric).

This subcase is impossible; otherwise, it gives rise to a cycle containing $y,x_{3},x_{4}\/$, a contradiction to the assumption (just consider in $G\/$ a path from $x_{1}\/$ to $u_{2}\/$ avoiding $u_{1}\/$).

It is straightforward to see that $x_{i}\notin N(x_{j})$ for $i=3,4$ and $j=1,2$ for all choices of $i$ and $j$; otherwise, $l_{i}>6$ or there exists a cycle containing $y,x_{3},x_{4}$. Therefore, subcase (b1) is finished.

(b2) Suppose $l_{3}=5$.

We may assume without loss of generality that $u_{3}=x_{1},\ x_{3}=u_{1}\/$ and $v_{3}=v_{2}\/$.

Suppose $d_{G}(v_{3})=2$. Consider $G^{\prime}=G-\{x_{3},v_{3}\}$; it is a non-trivial block chain with pendant edges $x_{1}v_{1},x_{2}u_{2}$. By Corollary 1(ii), there exists a hamiltonian path $P(x_{1},x_{2})\subseteq(G^{\prime})^{2}\/$ starting with $x_{1}v_{1}\/$ and ending with $u_{2}x_{2}\/$. We proceed block by block to construct $P(x_{1},x_{2})\/$ such that $x_{4}z_{4}\in E(G)\cap P(x_{1},x_{2})\/$ and $x_{4}z_{4}\not\in\{x_{1}v_{1},u_{2}x_{2}\}\/$: this is clear if $x_{4}\/$ is a cutvertex of $G^{\prime}\/$; and if $x_{4}\in V(B_{4})\/$ where $B_{4}\subseteq G^{\prime}\/$ is a $2\/$-connected block containing the cutvertices $c_{4},c_{4}^{\prime}\/$ of $G^{\prime}\/$, one uses a hamiltonian path $P(c_{4},c_{4}^{\prime})\/$ in $(B_{4})^{2}\/$ containing an edge incident to $x_{4}\/$ (Theorem F(i)). Then

[TABLE]

is a required hamiltonian path in $G^{2}$.

If $d_{G}(v_{3})>2$, then $G^{(0)}=G-\{x_{1},x_{2},x_{3}\}$ is connected (or else $v_{3}$ is a cutvertex of $G\/$). Any $v_{3}u_{2}\/$-path $P(v_{3},u_{2})\subset G^{(0)}\/$ can be extended to a cycle $yx_{1}x_{3}P(v_{3},u_{2})x_{2}y\/$ of length $\geq 6$, contradicting the assumption of this subcase.

(b3) Suppose $l_{3}=4\/$.

In this case, let $G^{\prime}=G-x_{3}$. Operating with $P(x_{1},x_{2})\subseteq(G^{\prime})^{2}$ as in case (b2), we obtain an ${\cal F}_{4}\/$ $x_{1}x_{2}\/$-hamiltonian path $(P(x_{1},x_{2})-u_{2}x_{2})u_{2}x_{3}x_{2}\/$ in $G^{2}$.

(1.4) $d_{G}(x_{3})=2,d_{G}(x_{4})>2$.

This case is symmetrical to the case (1.2).

(E)(2) Suppose $x_{3}$ and $x_{4}$ are in $K^{+}$.

Without loss of generality, assume that

[TABLE]

As for the definition of $x_{3}^{*},x_{4}^{*}$ see the paragraph preceding the statement of Lemma 2.

(2.1) $x_{3}\neq u_{1}$ and $x_{4}\neq u_{2}$.

(a) Suppose either $u_{i-2}\not\in N_{G}(x_{i})$, or $u_{i-2}\in\in N_{G}(x_{i})$ and $d_{G}(x_{i})>2$ for some $i\in\{3,4\}$. Without loss of generality, assume that $i=4$.

If $u_{1}\neq x_{3}^{*}$, set $W=\{y,u_{1},u_{2},x_{3}^{*},x_{4}^{*}\}$. Then $|W|=5$ and $K^{+}$ is $W$-sound, so by Theorem A, $G^{+}$ has a $W$-$EPS$-graph $S=E\cup P$ with $K^{+}\subseteq E$.

If $u_{1}=x_{3}^{*}$, then $d_{G}(x_{3})=2$ since $x_{3}\neq u_{1}$ by supposition. Now, let $S=E\cup P$ be an $[x_{4}^{*};u_{1},u_{2}]$-$EPS$-graph of $G^{+}$ with $K^{+}\subseteq E$ by Theorem C.

In either case, a required hamiltonian cycle in $(G^{+})^{2}$ can be constructed.

(b) Suppose $u_{i-2}\in N_{G}(x_{i})$ and $d_{G}(x_{i})=2$ for $i=3,4$.

If $w_{4}$ is the predecessor of $x_{4}$ in $K^{+}$ and $w_{4}\neq x_{3}$, then let $S=E\cup P$ be a $[x_{1};u_{1},u_{2},w_{4}]$-$EPS$-graph with $K^{+}\subseteq E$ by Theorem B. If $w_{4}=x_{3}$, then let $S=E\cup P$ be an $[x_{1};u_{1},u_{2}]$-$EPS$-graph with $K^{+}\subseteq E$ by Theorem C. Hence a required hamiltonian cycle in $(G^{+})^{2}$ can be constructed from $S$.

(2.2) $x_{3}=u_{1}$ and $x_{4}\neq u_{2}$.

(a) Suppose either $u_{2}\notin N_{G}(x_{4})$, or $u_{2}\in N_{G}(x_{4})$ and $d_{G}(x_{4})>2$.

(a1) $x_{3}x_{4}\in E(G)$.

If $d_{G}(x_{4})>2$, then $d_{G}(x_{3})=2$ and we choose an $[x_{1};x_{4},u_{2}]$-$EPS$-graph $S=E\cup P$ of $G^{+}$ with $K^{+}\subseteq E$ by Theorem C. If, however $d_{G}(x_{4})=2$, we choose an $[x_{1};x_{3},z_{4},u_{2}]$-$EPS$-graph $S=E\cup P$ of $G^{+}$ with $K^{+}\subseteq E$ by Theorem B. In either case, $S^{2}$ contains a required hamiltonian cycle.

(a2) $x_{3}x_{4}\notin E(G)$.

Here $w_{3}$ is the successor of $x_{3}$ in $K^{+}$. Let $S=E\cup P$ be an $[x_{3};u_{2},w_{3},x_{4}^{*}]$-$EPS$-graph with $K^{+}\subseteq E$ by Theorem B. Also here, $S^{2}$ contains a required hamiltonian cycle; it contains $x_{3}v\in E(G)$ which is consecutive to $x_{1}x_{3}$ in the eulerian trail of the component of $E$ containing $K^{+}$ (possibly $v=w_{3}$) and it contains $x_{4}z_{4}$.

(b) Suppose $u_{2}\in N(x_{4})$ and $d_{G}(x_{4})=2$.

(b1) $x_{3}x_{4}\in E(G)$.

Let $H=G-x_{4}\/$. Suppose $H\/$ is $2\/$-connected. Then by induction, $H\/$ has an ${\cal F}_{4}\/$ $x_{1}x_{2}\/$-hamiltonian path $P(x_{1},x_{2})\/$ in $H^{2}\/$ containing $x_{3}w_{3}\/$ and $u_{2}w_{2}\/$ which are edges of $H\/$. By deleting $u_{2}w_{2}\/$ from $P(x_{1},x_{2})\/$ and joining $x_{4}\/$ to $u_{2},w_{2}\/$, we obtain an ${\cal F}_{4}\/$ $x_{1}x_{2}\/$-hamiltonian path in $G^{2}\/$ containing $x_{3}w_{3},x_{4}u_{2}\/$ which are edges of $G\/$.

Suppose $H\/$ is not $2\/$-connected. Then $H\/$ is a non-trivial block chain with endblock $B_{i}\/$ containing $u_{i}\/$; $u_{i}\/$ is not a cutvetex of $H\/$, $i=1,2\/$. Let $c_{i}\/$ denote the cutvertex of $H\/$ which is contained in $B_{i}\/$, $i=1,2\/$. Set $B_{1,2}=H-(B_{1}\cup B_{2})\/$. If $c_{1}=c_{2}\/$, then set $B_{1,2}=c_{1}\/$. In any case, $c_{1}\/$ and $c_{2}\/$ are not cutvertices of $B_{1,2}\/$.

By supposing $x_{i}\neq c_{i}\/$ (and thus $B_{i}\/$ is $2\/$-connected) we apply Theorem F to conclude that $(B_{i})^{2}\/$ has an ${\cal F}_{3}\/$ $x_{i}c_{i}\/$-hamiltonian path $P(x_{i},c_{i})\/$, $i=1,2\/$ containing $x_{3}w_{3},u_{2}w_{2}\/$ respectively, which are edges of $G\/$. Let $P(c_{1},c_{2})\/$ denote a $c_{1}c_{2}\/$-hamiltonian path in $(B_{1,2})^{2}\/$. By deleting the edge $u_{2}w_{2}\/$ from the $x_{1}x_{2}\/$-hamiltonian path $P(x_{1},c_{1})P(c_{1},c_{2})P(x_{2},c_{2})\/$ in $(G-x_{4})^{2}\/$ and joining $x_{4}\/$ to $u_{2},w_{2}\/$, we obtain an ${\cal F}_{4}\/$ $x_{1}x_{2}\/$-hamiltonian path in $G^{2}\/$ containing $x_{3}w_{3},x_{4}u_{2}\/$ which are edges of $G$. Now suppose $x_{1}=c_{1}$ or $x_{2}=c_{2}$; i.e., $d_{G}(u_{1})=2$ or $d_{G}(u_{2})=2\/$. In this case we consider $G^{+}\/$ and choose an $[x_{1};u_{1},u_{2}]$-$EPS$-graph $S=E\cup P$ of $G^{+}$ with $K^{+}\subseteq E\/$ by Theorem C. Hence $S^{2}$ contains a hamiltonian cycle as required.

(b2) $x_{3}x_{4}\notin E(G)$.

If $w_{3}\neq w_{4}$, then we set $S=E\cup P$ to be an $[x_{3};u_{2},w_{3},w_{4}]$-$EPS$-graph of $G^{+}$ with $K^{+}\subseteq E$ by Theorem B. If $w_{3}=w_{4}$, then we set $S=E\cup P$ to be an $[x_{3};u_{2},w_{3}]$-$EPS$-graph of $G^{+}$ with $K^{+}\subseteq E$ by Theorem C. Here $w_{3}$ is the successor of $x_{3}$ and $w_{4}$ is the predecessor of $x_{4}$ in $K^{+}$. Hence $S^{2}$ yields a required hamiltonian cycle unless $w_{3}=w_{4}$ and $d_{G}(w_{3})>2$, in which case $d_{G}(x_{3})=2$ holds, and we operate with an $[x_{1};w_{3},u_{2}]$-$EPS$-graph by Theorem C. This settles case (2.2).

Since the case $x_{3}\neq u_{1}$ and $x_{4}=u_{2}$ is symmetrical to the case (2.2) just dealt with, we are left with the following case.

(2.3) $x_{3}=u_{1}$ and $x_{4}=u_{2}$.

(a) $d_{G}(x_{3})=2$.

(a1) $x_{3}x_{4}\notin E(G)$.

Choose an $[x_{4};u_{3},u_{4}]$-$EPS$-graph $S=E\cup P$ of $G^{+}$ with $K^{+}\subseteq E$ by Theorem C if $u_{3}\neq u_{4}$, and an $[x_{4};u_{3},u_{4}]$-$EPS$-graph $S=E\cup P$ of $G^{+}$ with $K^{+}\subseteq E$ by Theorem D if $u_{3}=u_{4}$; here $u_{3}$ is taken to be the successor of $x_{3}$ and $u_{4}$ the predecessor of $x_{4}$ in $K^{+}$. Then $S^{2}$ yields a required hamiltonian cycle unless $u_{3}=u_{4}$ and $d_{G}(u_{3})>2$. In this case $d_{G}(x_{4})=2$ and we may operate with an $[x_{2};u_{3}]$-$EPS$-graph to obtain a required hamiltonian cycle in $S^{2}$ by Theorem D.

(a2) $x_{3}x_{4}\in E(G)$.

(i) Suppose $d_{G}(x_{4})>2$.

$G-x_{3}\/$ is a block chain in which $x_{1}\/$ and $x_{4}\/$ are not cutvertices and belong to different endblocks. However, the endblock containing $x_{4}\/$ is $2\/$-connected since $d_{G}(x_{4})>2\/$; and it contains $x_{2}\/$ as well which is not a cutvertex of $G-x_{3}\/$ either. Therefore, $G^{+}-x_{3}$ is $2$-connected. Set

[TABLE]

$H\/$ is $2\/$-connected since $G^{+}-x_{3}\/$ is $2\/$-connected. By Theorem E, $H^{2}\/$ has a hamiltonian cycle $C\/$ containing $v_{1}x,xx_{2},x_{4}w_{4}\/$ which are edges of $H\/$. Now $(C-x)\cup\{v_{1}x_{3}x_{1}yx_{2}\}\/$ is a hamiltonian cycle in $(G^{+})^{2}\/$ with the required properties.

(ii) Suppose $d_{G}(x_{4})=2$.

Let $H\/$ be the graph obtained from $G^{+}\/$ by deleting $y,x_{2},x_{3},x_{4}\/$. Then $H\/$ is a non-trivial block chain containing $x_{1}\/$ which is not a cutvertex of $H\/$. By Corollary 1(i), $H^{2}\/$ has a hamiltonian cycle $C\/$ containing the edge $x_{1}v_{1}\/$ (which is an edge of $G\/$). This implies that the cycle $yx_{1}(C-x_{1}v_{1})v_{1}x_{3}x_{4}x_{2}y\/$ is a hamiltonian cycle in $(G^{+})^{2}\/$ having the required properties.

(b) $d_{G}(x_{3})>2$, hence $d_{G}(x_{4})>2$; otherwise we are back to (a) above, by symmetry. Then $x_{3}x_{4}\notin E(G)$.

Suppose $G^{\prime}=G-x_{1}$ is $2$-connected. Then by induction, $G^{\prime}$ has an ${\cal F}_{4}$ $v_{1}x_{2}$-hamiltonian path $P(v_{1},x_{2})$ in $(G^{\prime})^{2}$ containing $x_{3}w_{3}$ and $x_{4}w_{4}$ which are edges of $G^{\prime}$. Now $\{x_{1}v_{1}\}\cup P(v_{1},x_{2})$ is an ${\cal F}_{4}\/$ $x_{1}x_{2}$-hamiltonian path in $G^{2}$ containing $x_{3}w_{3},x_{4}w_{4}$ which are edges of $G$.

Now suppose $G^{\prime}=G-x_{1}$ is not $2$-connected. Then $G^{\prime}$ is a non-trivial block chain with $x_{3},v_{1}$ in different endblocks and not cutvertices. Note that the block containing $x_{3}$ is 2-connected and at least one block contaning $x_{4}$ is 2-connected, since $d_{G}(x_{3})>2$ and $d_{G}(x_{4})>2$.

(b1) Suppose $x_{2}$ is a cutvertex of $G^{\prime}$. Let $G_{1}$ and $G_{2}$ be the components of $G^{\prime}-x_{2}$ with either $x_{3},x_{4}\in V(G_{1})$ and $v_{2},v_{1}\in V(G_{2})$, or $x_{3},v_{2}\in V(G_{1})$ and $x_{4},v_{1}\in V(G_{2})$ (note $d_{G^{\prime}}(x_{2})=2$). Observe that in the first case $v_{2}=v_{1}$ is possible. However, $v_{1}=x_{4}$ is impossible because of the assumptions of this case (b); i.e., $d_{G}(x_{4})>2$. By the same token $v_{2}=x=3$ is impossible.

Suppose $x_{3},x_{4}\in V(G_{1})$ and $v_{2},v_{1}\in V(G_{2})$. Then by Theorem F(ii) or Corollary 1(ii), respectively, $(G_{1})^{2}$ has an $x_{3}x_{4}$-hamiltonian path $P_{1}$ containing an edge $x_{3}w_{3}\in E(G)$. If $G_{2}=K_{1}=v_{1}$, then we set $P=P_{1}\cup\{x_{2}x_{4},x_{3}v_{1},v_{1}x_{1}\}$. If $G_{2}=K_{2}=v_{2}v_{1}$, then we set $P=P_{1}\cup\{x_{2}x_{4},x_{3}v_{1},v_{1}v_{2},v_{2}x_{1}\}$. Otherwise, by Theorem E or Corollary 1(i), respectively, $(G_{2})^{2}$ has a hamiltonian cycle $C_{2}$ containing an edge $t_{1}v_{1}\in E(G)$. Then we set $P=P_{1}\cup C_{2}\cup\{x_{2}x_{4},x_{3}v_{1},t_{1}x_{1}\}-\{t_{1}v_{1}\}$. In all cases $P$ is an ${\cal F}_{4}$ $x_{1}x_{2}$-hamiltonian path in $G^{2}$ containing $x_{3}w_{3},x_{4}x_{2}$ which are edges of $G$ as required.

Suppose $x_{3},v_{2}\in V(G_{1})$ and $x_{4},v_{1}\in V(G_{2})$. Then we apply an analogous strategy as in the preceding case using Theorems E, F and Corollary 1, but considering $G_{1}$ instead of $G_{2}$ and vice versa.

(b2) Suppose $x_{2}$ is not a cutvertex of $G^{\prime}$. Let $B_{2}$ be the 2-connected block containing $x_{2}$.

(i) Suppose $x_{3}\in V(B_{2})$. Let $t$ be the cutvertex of $G^{\prime}$ in $B_{2}$; possibly $t=x_{4}$, $t\notin\{x_{2},x_{3}\}$ in any case. We define the block chain $G_{1}$ such that $G^{\prime}=B_{2}\cup G_{1}$ and $B_{2}\cap G_{1}=\{t\}$. If $t=x_{4}$, then $(B_{2})^{2}$ has an $x_{2}t$-hamiltonian path $P_{2}$ containing $x_{3}w_{3}\in E(G)$ by Theorem F(i). If $t\neq x_{4}$, then by induction $(B_{2})^{2}$ has an $x_{2}t$-hamiltonian path $P_{2}$ containing $x_{3}w_{3},x_{4}w_{4}$ which are different edges of $G$. In both cases $(G_{1})^{2}$ has a $tv_{1}$-hamiltonian path starting with $tw\in E(G)$, by Theorem F(ii) or Corollary 1(ii), respectively. Then $P=P_{2}\cup P_{1}\cup\{v_{1}x_{1}\}$ is an ${\cal F}_{4}$ $x_{1}x_{2}$-hamiltonian path in $G^{2}$ containing $x_{3}w_{3},x_{4}w_{4}$ which are edges of $G$ as required. Note that if $t=x_{4}$, then $x_{4}w_{4}=tw$.

(ii) Suppose $x_{3}\notin V(B_{2})$. If $B_{2}$ is not an endblock, then $t,t^{\prime}$ denote the cutvertices of $G^{\prime}$ in $B_{2}$ and we define block chains $G_{0}$, $G_{1}$ such that $G^{\prime}=G_{1}\cup B_{2}\cup G_{0}$, $x_{3}\in V(G_{1}),v_{1}\in V(G_{0})$ and $G_{1}\cap B_{2}=t$, $B_{2}\cap G_{0}=t^{\prime}$. If $B_{2}$ is an endblock, then we proceed analogously: we set $G_{0}=\emptyset$ and $t^{\prime}=v_{1}$ in this case. Note that $t=x_{4}$ ot $t^{\prime}=x_{4}$ is possible.

If $t^{\prime}\neq x_{4}$, then by Theorem F(i) $(B_{2})^{2}$ has an $x_{2}t$-hamiltonian path $P_{2}$ containing $t^{\prime}w^{\prime}\in E(G)$ for $t=x_{4}$ and by induction $(B_{2})^{2}$ has an ${\cal F}_{4}$ $x_{2}t$-hamiltonian path $P_{2}$ containing $t^{\prime}w^{\prime},x_{4}w_{4}$ which are different edges of $G$ for $t\neq x_{4}$. By the same token $(G_{1})^{2}$ has an $tx_{3}$-hamiltonian path $P_{1}$ containing $tw\in E(G)$. If $G_{0}=\emptyset$, then we set $P=P_{2}\cup P_{1}\cup\{x_{3}x_{1}\}$. If $G_{0}=t^{\prime}v_{1}$, then we set $P=P_{2}\cup P_{1}\cup\{x_{3}x_{1},w^{\prime}v_{1},v_{1}t^{\prime}\}-\{t^{\prime}w^{\prime}\}$. Otherwise $(G_{0})^{2}$ has a hamiltonian cycle $C_{0}$ containing $t^{\prime}w^{*}\in E(G)$ by Theorem E or Corollary 1(i), respectively, and we set $P=P_{2}\cup C_{0}\cup P_{1}\cup\{x_{3}x_{1},w^{\prime}w^{*}\}-\{t^{\prime}w^{\prime},t^{\prime}w^{*}\}$. In all cases $P$ is an ${\cal F}_{4}$ $x_{1}x_{2}$-hamiltonian path in $G^{2}$ containing $x_{3}x_{1},x_{4}w_{4}$ which are edges of $G$ as required. Note that if $t=x_{4}$, then $x_{4}w_{4}=tw$.

If $t^{\prime}=x_{4}$, we proceed analogously as in the previous case with $G_{1}$ and $G_{0}$ switching roles.

This completes the proof of Theorem 2.

Acknowledgements

Research of the first author was supported by the FRGS Grant (FP036-2013B), the second author was supported by project P202/12/G061 of the Grant Agency of the Czech Republic, whereas research of the third author was supported in part by FWF-grant P27615-N25.

This publication was partly supported by the project LO1506 of the Czech Ministry of Education, Youth and Sports.