Hamiltonian Laceability of Hypercubes Without Isometric Subgraphs

Abstract

Locke (Am Math Mon 108:668, 2001) conjectured that the n-dimensional hypercube $Q_n$ with the set F of 2k removed vertices half from each bipartition set, where $n \ge k + 2$ and $k \ge 1$, is Hamiltonian. So far the conjecture remains open although partial results are known; some of them with additional conditions on the set F. We explore Hamiltonian properties of $Q_n - F$ if the set of faulty vertices F forms either an isometric cycle of $Q_n$ or an isometric tree of $Q_n$. We study a more general problem. A bipartite graph G is Hamiltonian laceable if either (a) its bipartition sets are of equal size and for each pair of vertices x, y from different bipartition sets there exists a Hamiltonian path between x and y, or (b) its bipartition sets differ in sizes by one and for each pair of vertices x, y from the larger bipartition set there exists a Hamiltonian path between x and y. In particular, we show that if C is an isometric cycle in $Q_n$ for $n \ge 5$, then $Q_n - V(C)$ is Hamiltonian laceable. This allows us to remove up to 2n faulty vertices. Furthermore, if T is balanced isometric tree in $Q_n$, then for $n \ge 4$ the graph $Q_n - V(T)$ is Hamiltonian laceable. Finally, if T is an almost-balanced isometric tree in $Q_n$, then for $n \ge 5$ the graph $Q_n - V(T)$ is Hamiltonian laceable. Thus our results support Locke hypothesis.

Appendices

Appendix A: An Isometric Cycle of Length 10 in $Q_5$

We show that Theorem 2 holds for $n = 5$. That is, if C is an isometric cycle in $Q_5$, then $Q_5 - V(C)$ is Hamiltonian laceable, see Theorem 5. Its proof is a tedious case analysis, so we decided to put it in the appendix. We start by considering an isometric path $P_4$ of length 4 in $Q_4$. It is easy to observe that $Q_4 - V(P_4)$ is not Hamiltonian laceable.

Fig. 32

A hypercube $Q_4 - V(P_4)$ used in Lemmas 10 and 11

Remark 7

Let $P_4$ be an isometric path in $Q_4$ and let $c_4$ denote the only black vertex of degree two in $Q_4 - V(P_4)$. We denote the only two neighbors of $c_4$ in $Q_4 - V(P_4)$ by $u_5$ and $u_6$, see Fig. 32. Then there does not exist a Hamiltonian path between $u_5$ and $u_6$ in $Q_4 - V(P_4)$. We show this by contradiction; we assume that such path exists and we denote it by H. The path H contains the vertex $c_4$. Since the only neighbors of $c_4$ are the vertices $u_5$ and $u_6$, it must be $H = (u_5, c_4, u_6)$ which is clearly not a Hamiltonian path in $Q_4 - V(P_4)$.

Lemma 10 proves that this counterexample to Hamiltonian laceability of $Q_4 - V(P_4)$ is unique. Lemma 11 shows that for every two vertices b, w in $Q_4 - V(P_4)$ of opposite parity there exists a Hamiltonian path between b and w in $Q_4$ passing through all the edges of $P_4$. Moreover, we show that this Hamiltonian path avoids certain vertices. Both lemmas are proved by presenting a list of Hamiltonian paths between all desired vertices. Since $Q_4$ is a small graph, we believe that this kind of proof is shorter and more readable than an exhausting case analysis.

Lemma 10

Let $P_4$ be an isometric path of length 4 in $Q_4$ and let $u_5, u_6 \in W_4$ be the vertices in Fig. 32. Then for every distinct $w, w' \in W_4 {\setminus } V(P_4)$ except the case ${\left\{ w, w' \right\} } = {\left\{ u_5, u_6 \right\} }$ there exists a Hamiltonian path between w and $w'$ in $Q_4 - V(P_4)$.

Proof

We denote the white vertices of $Q_4 - V(P_4)$ by $u_1, u_2, u_3, u_4, u_5, u_6$, the black vertices of $Q_4 - V(P_4)$ by $c_1, c_2, c_3, c_4, c_5$ and the vertices of $P_4$ by $P_4 = (b_0, w_1, b_2, w_3, b_4)$, see Fig. 32. We present all 14 Hamiltonian paths in $Q_4 - V(P_4)$ between $u_i$ and $u_j$ for every $1 \le i < j \le 6$ except the case $\{u_i, u_j\} = \{u_5, u_6\}$:

$\square$

Lemma 11

Let $P_4 = (b_0, w_1, b_2, w_3, b_4)$ be an isometric path of length 4 in $Q_4$ and let $u_2, u_3 \in W_4$ be the vertices in Fig. 32. Then for every $b \in B_4 {\setminus } V(P_4)$ and every $w \in W_4 {\setminus } V(P_4)$ there exist vertices $x, y \in W_4 {\setminus } V(P_4)$ such that x is a neighbor of $b_0$ and y is a neighbor of $b_4,$ and there exist paths $H_1 = (b H_1 x)$ and $H_2 = (y H_2 w)$ in $Q_4 - V(P_4)$ such that $(b H_1 x, b_0 P_4 b_4, y H_2 w)$ is a Hamiltonian path between b and w in $Q_4$ passing through all the edges of $P_4$.

Proof

We denote the white vertices of $Q_4 - V(P_4)$ by $u_1, u_2, u_3, u_4, u_5, u_6$ and we denote the black vertices of $Q_4 - V(P_4)$ by $c_1, c_2, c_3, c_4, c_5$, see Fig. 32. We present all 30 Hamiltonian paths in $Q_4$ between $c_i$ and $u_j$ for every $1 \le i \le 5$ and for every $1 \le j \le 6$ passing through all the edges of $P_4$:

$\square$

Fig. 33

The graph $Q_5 - V(C)$ with edges of direction 1 to 5

The following three lemmas separately consider three cases in the proof of Theorem 5.

Lemma 12

Let C be an isometric cycle of length 10 in $Q_5$ and let $i \in {\left\{ 1,2,3,4,5 \right\} }$ be a direction of any edge $st \in E(C)$ where $s \in B_5$. Then for every $b \in B_5 {\setminus } V(C)$ that has the same ith coordinate as t and for every $w \in W_5 {\setminus } V(C)$ that has the same ith coordinate as s there exists a Hamiltonian path between b and w in $Q_5 - V(C)$.

Proof

We denote the vertices of C by $C = (b_0,$ $w_1,$ $b_2,$ $w_3,$ $b_4,$ $w_5,$ $b_6,$ $w_7,$ $b_8, w_9)$, every $b_i$ is in $B_5$ and every $w_i$ is in $W_5$. For simplicity of the notation we assume that i is the direction of the edge $b_0w_1$, so $s = b_0$ and $t = w_1$. Let $P^0 = (b_6, w_7, b_8, w_9, b_0)$ and $P^1 = (w_1, b_2, w_3, b_4, w_5)$ be the subpaths of C. Since $Q_5$ is vertex-transitive we can assume that $b_0$ is the zero vertex. The direction i splits $Q_5$ into $Q_4^{0}$ containing the path $P^0$ and $Q_4^{1}$ containing the path $P^1$. We denote the white vertices of $Q_{4}^{0} - V(P^0)$ by $u_1, u_2, u_3, u_4, u_5, u_6$ and the black vertices of $Q_{4}^{1} - V(P^1)$ by $d_1, d_2, d_3, d_4, d_5, d_6$ as in Fig. 33. Note that there is no Hamiltonian path between $u_5$ and $u_6$ in $Q_4^{0} - V(P^0)$ and there is no Hamiltonian path between $d_2$ and $d_3$ in $Q_4^{1} - V(P^1)$, see Remark 7. There are three cases to consider regarding the vertices b and w.

Case 1: The vertex b is not in ${\left\{ d_2, d_3 \right\} }$. We choose an edge uv of direction i such that $u \in {\left\{ u_2, u_3 \right\} }$ and $u \ne w$. By Lemma 10, there exists a Hamiltonian path $H_1$ between w and u in $Q_{4}^{0} - V(P^0)$ since $u \notin {\left\{ u_5, u_6 \right\} }$. By Lemma 10, there exists a Hamiltonian path $H_2$ between b and v in $Q_{4}^{1} - V(P^1)$ since $b \notin {\left\{ d_2, d_3 \right\} }$. By joining $H_1$ and $H_2$ we obtain a Hamiltonian path $(b H_2 v, u, H_1 w)$ in $Q_5 - V(C)$.

Case 2: The vertex b is in ${\left\{ d_2, d_3 \right\} }$ and the vertex w is not in ${\left\{ u_5, u_6 \right\} }$. We choose an edge uv of direction i such that $v \in {\left\{ d_5, d_6 \right\} }$ and $u \ne w$. By Lemma 10, there exists a Hamiltonian path $H_1$ between w and u in $Q_{4}^{0} - V(P^0)$ since $w \notin {\left\{ u_5, u_6 \right\} }$. By Lemma 10, there exists a Hamiltonian path $H_2$ between b and v in $Q_{4}^{1} - V(P^1)$ since $v \notin {\left\{ d_2, d_3 \right\} }$. By joining the paths $H_1$ and $H_2$ we obtain a Hamiltonian path $(b H_2 v, u H_1 w)$ in $Q_5 - V(C)$.

Case 3: The vertex b is in ${\left\{ d_2, d_3 \right\} }$ and the vertex w is in ${\left\{ u_5, u_6 \right\} }$. We show an existence of four Hamiltonian paths in $Q_5 - V(C)$. First we define four auxiliary paths:

Paths $H_1 = (d_2, v_3, d_3, u_3, c_3, u_4, c_2, u_2, c_1, u_1, c_5)$ and $H_2 = (u_5, c_4, v_5, d_5, v_1, d_1, v_2, d_4, v_4, d_6, u_6)$, see Fig. 34.

Paths $H_3 = (d_3, v_3, d_2, u_2, c_1, u_1, c_2, u_3, c_3, u_4, c_5)$ and $H_4 = (u_5, d_5, v_1, d_1, v_2, d_4, v_4, d_6, v_5, c_4, u_6)$, see Fig. 35.

We combine these paths to obtain four desired Hamiltonian paths in $Q_5 - V(C)$: $(d_2 H_1 c_5, u_5 H_2 u_6)$, $(d_2 H_1 c_5, u_6 H_4 u_5)$, $(d_3 H_3 c_5, u_5 H_2 u_6)$, and $(d_3 H_3 c_5, u_6 H_4 u_5)$.

$\square$

Fig. 34

The graph $Q_5 - V(C)$ with paths $H_1$ and $H_2$

Fig. 35

The graph $Q_5 - V(C)$ with paths $H_3$ and $H_4$

Lemma 13

Let C be an isometric cycle of length 10 in $Q_5$ and let $i \in {\left\{ 1,2,3,4,5 \right\} }$ be a direction of an edge $st \in E(C)$ where $s \in B_5$. Then for every $b \in B_5 {\setminus } V(C)$ and for every $w \in W_5 {\setminus } V(C)$ that both have the same ith coordinate as s there exists a Hamiltonian path between b and w in $Q_5 - V(C)$.

Proof

We denote the vertices of C by $C = (b_0,$ $w_1,$ $b_2,$ $w_3,$ $b_4,$ $w_5,$ $b_6,$ $w_7,$ $b_8, w_9)$ so that every $b_i$ is in $B_5$ and every $w_i$ is in $W_5$. For simplicity of the notation we assume that i is the direction of the edge $b_0w_1$, so $s = b_0$ and $t = w_1$. Let $P^0 = (b_6, w_7, b_8, w_9, b_0)$ and $P^1 = (w_1, b_2, w_3, b_4, w_5)$ be the subpaths of C. Since $Q_5$ is vertex-transitive, we can assume that $b_0$ is the zero vertex. The direction i splits $Q_5$ into $Q_4^{0}$ containing the path $P^0$ and $Q_4^{1}$ containing the path $P^1$. We denote the white vertices of $Q_4^0 - V(P^0)$ by $u_1, u_2, u_3, u_4, u_5, u_6$ and we denote the black vertices of $Q_4^1 - V(P^1)$ by $d_1, d_2, d_3, d_4, d_5$ as in Fig. 33.

By Lemma 11, there exist vertices $x \in {\left\{ u_2, u_4, u_5 \right\} }$ and $y \in {\left\{ u_1, u_3, u_6 \right\} }$ such that ${\left\{ x, y \right\} } \ne {\left\{ u_2, u_3 \right\} }$ and there exist paths $H_1 = (b H_1 x)$ and $H_2 = (y H_2 w)$ in $Q_4^0 - V(P^0)$ so that $(b H_1 x, b_0 P_4 b_4, y H_2 w)$ is a Hamiltonian path between b and w in $Q_4$ passing through all the edges of $P_4$. Let d and $d'$ denote the neighbors of x and y in $Q_4^1$, respectively. Note that there does not exist a Hamiltonian path between $d_2$ and $d_3$ in $Q_4^{1} - V(P^1)$, see Remark 7. By Lemma 10, there exists a Hamiltonian path $H_3$ between d and $d'$ in $Q_{4}^{1} - V(P^1)$ since ${\left\{ d, d' \right\} } \ne {\left\{ d_2, d_3 \right\} }$. We join the paths $H_1$, $H_2$ and $H_3$ to obtain a Hamiltonian path $(b H_1 x, d H_3 d', y H_2 w)$ between b and w in $Q_4 - V(C)$.$\square$

Lemma 14

Let $C = (b_0,$ $w_1,$ $b_2,$ $w_3,$ $b_4,$ $w_5,$ $b_6,$ $w_7,$ $b_8, w_9)$ be an isometric cycle in $Q_5,$ every $b_i$ is in $B_5$ and every $w_i$ is in $W_5$. Let i be the direction of the edge $b_0w_1$. Then for every $b \in B_5 {\setminus } V(C)$ that has the same ith coordinate as $b_0$ and for every $w \in W_5 {\setminus } V(C)$ that has the same ith coordinate as $w_1$ there exists a Hamiltonian path between b and w in $Q_5 - V(C)$.

Proof

The direction i splits $Q_5$ into two subcubes $Q_{4}^{0,i}$ and $Q_{4}^{1,i}$. There are five black vertices in $Q_4^{0,i} - {\left\{ b_0, b_6, b_8 \right\} }$ which we label $c_1, c_2, c_3, c_4, c_5$ as in Fig. 33 and we denote the set of those vertices by U. There are five white vertices in $Q_4^{1,i} - {\left\{ w_1, w_3, w_5 \right\} }$ which we label $v_1, v_2, v_3, v_4, v_5$ as in Fig. 33 and we denote the set of those vertices by V. We say that H is a Hamiltonian path for a set ${\left\{ x, y \right\} } \subseteq {V(G)}$ in a graph G if (xH y) is a Hamiltonian path in G. We show an existence of all Hamiltonian paths between every $c \in U$ and every $v \in V$ in $Q_5 - V(C)$, which is 25 Hamiltonian paths.

For the following sets of vertices: $r_{1,1} = {\left\{ c_1, v_1 \right\} }$, $r_{1,2} = {\left\{ c_1, v_2 \right\} }$, $r_{1,5} = {\left\{ c_1, v_5 \right\} }$, $r_{2,3} = {\left\{ c_2, v_3 \right\} }$, $r_{2,4} = {\left\{ c_2, v_4 \right\} }$, $r_{3,3} = {\left\{ c_3, v_3 \right\} }$, $r_{3,4} = {\left\{ c_3, v_4 \right\} }$, $r_{4,1} = {\left\{ c_4, v_1 \right\} }$, $r_{4,2} = {\left\{ c_4, v_2 \right\} }$, $r_{4,5} = {\left\{ c_4, v_5 \right\} }$, $r_{5,3} = {\left\{ c_5, v_3 \right\} }$, $r_{5,4} = {\left\{ c_5, v_4 \right\} }$ we use the direction $j = 1$ of the edge $b_6w_7$ to split $Q_5$ into $Q_4^{0,1}$ containing $b_6$ and $Q_4^{1,1}$ containing $w_7$. The vertices of the set ${\left\{ c_2, c_3, c_5, v_3, v_4 \right\} }$ are in $Q_4^{0,1}$ and the vertices of the set ${\left\{ c_1, c_4, v_1, v_2, v_5 \right\} }$ are in $Q_4^{1,1}$. We use Lemma 13 for $i = 1$ to find Hamiltonian paths in $Q_5 - V(C)$ for the following sets: $r_{1,1}$, $r_{1,2}$, $r_{1,5}$, $r_{2,3}$, $r_{2,4}$, $r_{3,3}$, $r_{3,4}$, $r_{4,1}$, $r_{4,2}$, $r_{4,5}$, $r_{5,3}$, and $r_{5,4}$. This shows an existence of 12 Hamiltonian paths in $Q_5 - V(C)$.

For the following sets of vertices: $r_{1,3} = {\left\{ c_1, v_3 \right\} }$, $r_{1,4} = {\left\{ c_1, v_4 \right\} }$, $r_{2,1} = {\left\{ c_2, v_1 \right\} }$, $r_{2,2} = {\left\{ c_2, v_2 \right\} }$, $r_{3,1} = {\left\{ c_3, v_1 \right\} }$, $r_{3,2} = {\left\{ c_3, v_2 \right\} }$, $r_{3,5} = {\left\{ c_3, v_5 \right\} }$, $r_{4,4} = {\left\{ c_4, v_4 \right\} }$, $r_{5,1} = {\left\{ c_5, v_1 \right\} }$, $r_{5,2} = {\left\{ c_5, v_2 \right\} }$, $r_{5,5} = {\left\{ c_5, v_5 \right\} }$ we use the direction $j = 3$ of the edge $b_8w_9$ to split $Q_5$ into $Q_4^{0,3}$ containing $b_8$ and $Q_4^{1,3}$ containing $w_9$.

Fig. 36

The Hamiltonian path between $c_2$ and $v_5$ in $Q_5 - V(C)$

Fig. 37

The Hamiltonian path between $c_4$ and $v_3$ in $Q_5 - V(C)$

Table 1 Table showing an existence of all Hamiltonian paths between every $c \in U$ and every $v \in V$ in $Q_5 - V(C)$

The vertices of the set ${\left\{ c_1, v_1, v_2 \right\} }$ are in $Q_4^{0,3}$ and the vertices of the set ${\left\{ c_3, c_5, v_4 \right\} }$ are in $Q_4^{1,3}$. We use Lemma 12 for $i = 3$ to find Hamiltonian paths in $Q_5 - V(C)$ for the following sets: $r_{1,4}$, $r_{3,1}$, $r_{3,2}$, $r_{5,1}$, and $r_{5,2}$. This shows an existence of 5 Hamiltonian paths in $Q_5 - V(C)$.

The vertices of the set ${\left\{ c_1, c_2, v_1, v_2, v_3 \right\} }$ are in $Q_4^{0,3}$ and the vertices of the set ${\left\{ c_3, c_4, c_5, v_4, v_5 \right\} }$ are in $Q_4^{1,3}$. We use Lemma 13 for $i = 3$ to find Hamiltonian paths in $Q_5 - V(C)$ for the following sets: $r_{1,3}$, $r_{2,1}$, $r_{2,2}$, $r_{3,5}$, $r_{4,4}$, and $r_{5,5}$. This shows an existence of 6 Hamiltonian paths in $Q_5 - V(C)$.

We showed an existence of 23 Hamiltonian paths in $Q_5 - V(C)$. The remaining two Hamiltonian paths between $c_2$ and $v_5$ and between $c_4$ and $v_3$ are depicted on Figs. 36 and 37, respectively. We present an existence of all Hamiltonian paths in a form of Table 1. $\square$

Theorem 5

Let C be an isometric cycle in $Q_5$ of length 2k where $k \ge 3$ is odd. Then $Q_5 - V(C)$ is Hamiltonian laceable.

Proof

If $k = 3$ we use Proposition 4 since $2k \le 2 \cdot 5 -4$. From now on let $k = 5$. Let us denote the vertices of C by $C = (b_0,$ $w_1,$ $b_2,$ $w_3,$ $b_4,$ $w_5,$ $b_6,$ $w_7,$ $b_8, w_9)$ so that every $b_i$ is in $B_5$ and every $w_i$ is in $W_5$. Let $b \in B_5 {\setminus } V(C)$ and $w \in W_5 {\setminus } V(C)$. We show that there exists a Hamiltonian path between b and w in $Q_5 - V(C)$. Since $Q_5$ is vertex-transitive we can assume that $b_0$ is the zero vertex. We denote the direction of the edge $b_0w_1$ by i. Let $P^0 = (b_6, w_7, b_8, w_9, b_0)$ and $P^1 = (w_1, b_2, w_3, b_4, w_5)$ be the subpaths of C. Then $P^0$ is in $Q_{4}^{0,i}$ and $P^1$ is in $Q_{4}^{1,i}$. There are three cases to consider regarding the position of b and w.

Fig. 38

Hamiltonian paths (thick green) between $y \in Y$ and z in $Q_4 - (V(P_1) \cup V(e))$ for every vertex $y \in Y$ and every edge $e \in F$ (dashed blue) where $Y = {\left\{ y_1, y_2, y_3 \right\} } \subseteq V(Q_4)$ and $F = {\left\{ e_1, e_2, e_3, e_4, e_5, e_6 \right\} } \subseteq E(Q_4)$. All cases in Lemma 4 (color figure online)

Case 1: The vertex w is in $Q_{4}^{0,i}$ and the vertex b is in $Q_{4}^{1,i}$. This case is proved by Lemma 12.

Case 2: The vertices b, w are in the same subcube; say $Q_{4}^{0,i}$. This case is proved by Lemma 13.

Case 3: The vertex b is in $Q_{4}^{0,i}$ and the vertex w is in $Q_{4}^{1,i}$. This case is proved by Lemma 14. $\square$

Appendix B: The Proof of Lemma 4

Lemma 4 is proved by presenting all eighteen Hamiltonian paths, see Fig. 38. Here we repeat the statement of Lemma 4.

Lemma

Let $Y = {\left\{ y_1, y_2, y_3 \right\} } \subseteq V(Q_4),$ $F = {\left\{ e_1, e_2, e_3, e_4, e_5, e_6 \right\} } \subseteq E(Q_4)$ and $z, b_4, w_3, b_5 \in V(Q_4)$ as in Fig. 17. Then for every $y \in Y$ and every $e \in F$ there exists a Hamiltonian path between y and z in $Q_4 - ({\left\{ b_4, w_3, b_5 \right\} } \cup V(e))$.