假設G=(V_0 union V_1)是一個雙分圖，其中V_0和V_1代表的是G中獨立的點集合，且兩者大小相同，而E代表的是G中邊的集合。首先令x和y分別是圖中相異的任意兩點，而w是相異於 x和y的點，則我們若稱此G為”1-vertex-Hamiltonian-laceable ”就必須在 之情況下滿足以下三個性質：
性質一：x和y在同一個點集合中，且w在另一個點集合中
之情況下，x和y之間存在一條最長路徑長度為
(lV_0l+lV_1l-2)。
性質二：x和y在不同的點集合中，且w在任意的一個點集合
中之情況下，x和y之間存在一條最長路徑長度為
(lV_0l+lV_1l-3)。
性質三：x，y和w均在同一個點集合中，此時x和y之間存在
一條最長路徑長度為(lV_0l+lV_1l-4)。
在此篇論文中，我們將主要證明：當lF_el less than n-3時，Q_n - F_e仍然是1-vertex-Hamiltonian-laceable，而F_e代表的是n維超立方體中之錯邊集合。

Suppose that G = (V_0 union V_1,E) is a bipartite graph with two partite sets V_0 and V_1 of
equal size. Let x and y be two arbitrary distinct vertices and let w be another vertex different from x and y. G is said to be 1-vertex-Hamiltonian-laceable if G-w satisfies the following three properties.
P1: There is a (lV_0l + lV_1l - 2)-length path between x and y, where x and y are in the same partite set and w is in the other partite set;
P2: There is a (lV_0l + lV_1l - 3)-length path between x and y, where x and y are in different partite sets and w is in any partite set;
P3: There is a (lV_0l + lV_1l - 4)-length path between x and y, where x,y,w are in the same partite set.
Let Fe be the set of faulty edges of an n-dimensional hypercube Qn. In this paper, we show that Qn - Fe (the graph obtained by deleting all edges of Fe from Qn) remains
1-vertex-Hamiltonian-laceable when lFel less than n-3.

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