Let G = (V, E) be a planar triangulated graph (PTG) having every face triangular. A rectilinear dual or an orthogonal floor plan (OFP) of G is obtained by partitioning a rectangle into \mid V \mid rectilinear regions (modules) where two modules are adjacent if and only if there is an edge between the corresponding vertices in G. In this paper, a linear-time algorithm is presented for constructing an OFP for a given G such that the obtained OFP has B_{min} bends, where a bend in a concave corner in an OFP. Further, it has been proved that at least B_{min} bends are required to construct an OFP for G, where \rho - 2 \leq B_{min} \leq \rho + 1 and \rho is the sum of the number of leaves of the containment tree of G and the number of K_4 (4-vertex complete graph) in G.

中文翻译：

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一种用于构建具有最少弯曲次数的正交平面图的线性时间算法

让 G = (V, E) 是一个平面三角图 (PTG)，每个面都是三角形的。G 的直线对偶或正交平面图 (OFP) 是通过将矩形划分为 \mid V \mid 直线区域（模块）而获得的，其中两个模块相邻当且仅当 G 中的相应顶点之间存在边。在本文中，提出了一种线性时间算法，用于为给定的 G 构造一个 OFP，使得所获得的 OFP 具有 B_{min} 弯曲，其中在一个 OFP 的凹角处有一个弯曲。此外，已经证明至少需要 B_{min} 个弯曲来构造 G 的 OFP，其中 \rho - 2 \leq B_{min} \leq \rho + 1 和 \rho 是G 的包含树的叶子和 G 中 K_4（4 顶点完全图）的数量。